1.6: Modeling with Linear Functions - Mathematics

When modeling scenarios with a linear function and solving problems involving quantities changing linearly, we typically follow the same problem solving strategies that we would use for any type of function:

Problem Solving Strategy

  1. Identify changing quantities, and then carefully and clearly define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
  2. Carefully read the problem to identify important information. Look for information giving values for the variables, or values for parts of the functional model, like slope and initial value.
  3. Carefully read the problem to identify what we are trying to find, identify, solve, or interpret.
  4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table or even finding a formula for the function being used to model the problem.
  5. When needed, find a formula for the function.
  6. Solve or evaluate using the formula you found for the desired quantities.
  7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
  8. Clearly convey your result using appropriate units, and answer in full sentences when appropriate.

Example (PageIndex{1})

A company purchased $120,000 in new office equipment. Then expect the value to depreciate (decrease) by $16,000 per year. Find a linear model for the value, then find and interpret the horizontal intercept and determine a reasonable domain and range for this function.


In the problem, there are two changing quantities: time and value. The remaining value of the equipment depends on how long the company has had it. We can define our variables, including units.

Output: (V), value remaining, in dollars

Input: (t), time, in years

Reading the problem, we identify two important values. The first, $120,000, is the initial value for (V). The other value appears to be a rate of change – the units of dollars per year match the units of our output variable divided by our input variable. The value is depreciating, so you should recognize that the value remaining is decreasing each year and the slope is negative.

Using the intercept and slope provided in the problem, we can write the equation: (V(t) = 120,000 - 16,000t).

To find the horizontal intercept, we set the output to zero, and solve for the input:

[egin{align*} 0 &= 120,000 - 16,000t t &= frac{120,000}{16,000} = 7.5 end{align*} ]

The horizontal intercept is 7.5 years. Since this represents the input value where the output will be zero, interpreting this, we could say: The equipment will have no remaining value after 7.5 years.

When modeling any real life scenario with functions, there is typically a limited domain over which that model will be valid – almost no trend continues indefinitely. In this case, it certainly doesn’t make sense to talk about input values less than zero. This model is also not valid after the horizontal intercept.

The domain represents the set of input values and so the reasonable domain for this function is (0 leq t leqs 7.5). The range represents the set of output values and the value starts at $120,000 and ends with $0 after 7.5 years so the corresponding range is (0 leq V(t) leqs 120,000).

Most importantly remember that domain and range are tied together, and whatever you decide is most appropriate for the domain (the independent variable) will dictate the requirements for the range (the dependent variable).

Example (PageIndex{2})

Jamal is choosing between two moving companies. The first, U-Haul, charges an up-front fee of $20, then 59 cents a mile. The second, Budget, charges an up-front fee of $16, then 63 cents a mile(Rates retrieved Aug 2, 2010 from and When will U-Haul be the better choice for Jamal?


The two important quantities in this problem are the cost, and the number of miles that are driven. Since we have two companies to consider, we will define two functions:

Input: (m), miles driven


(Y(m)): cost, in dollars, for renting from U-Haul

(B(m)): cost, in dollars, for renting from Budget

Reading the problem carefully, it appears that we were given an initial cost and a rate of change for each company. Since our outputs are measured in dollars but the costs per mile given in the problem are in cents, we will need to convert these quantities to match our desired units: $0.59 a mile for U-Haul, and $0.63 a mile for Budget.

Looking to what we’re trying to find, we want to know when U-Haul will be the better choice. Since all we have to make that decision from is the costs, we are looking for when U-Haul will cost less, or when (Y(m) < B(m)). The solution pathway will lead us to find the equations for the two functions, find the intersection, then look to see where the (Y(m)) function is smaller. Using the rates of change and initial charges, we can write the equations:

[Y(m) = 20 + 0.59m onumber ]
[B(m) = 16 + 0.63m onumber ]

These graphs are sketched to the right, with Y(m) drawn dashed.

To find the intersection, we set the equations equal and solve:

[egin{array} {rcl} {Y(m)} &= & {B(m)} {20 + 0.59m} &= & {16 + 0.63m} {4} &= & {0.04m} {m} &= & {100} end{array} onumber ]

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that (Y(m)) is growing at a slower rate, we can conclude that U-Haul will be the cheaper price when more than 100 miles are driven.

Example (PageIndex{3})

A company’s revenue has been growing linearly. In 2012 their revenue was $1.45 million. By 2015 the revenue had grown to $1.81 million. If this trend continues,

a. Predict the revenue in 2020.

b. When will revenue reach $3 million?


The two changing quantities are the revenue and time. While we could use the actual year value as the input quantity, doing so tends to lead to very ugly equations, since the vertical intercept would correspond to the year 0, more than 2000 years ago!

To make things a little nicer, and to make our lives easier too, we will define our input as years since 2012:

Input: (t), years since 2012

Output: (R(t)), the company’s revenue, in millions of dollars

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2012 would correspond to (t = 0), giving the point ((0, 1.45)). Notice that through our clever choice of variable definition, we have “given” ourselves the vertical intercept of the function. The year 2015 would correspond to (t = 3), giving the point ((3, 1.81)).

To predict the population in 2020 ((t = 8)), we would need an equation for the population. Likewise, to find when the revenue would reach $3 million, we would need to solve for the input that would provide an output of 3. Either way, we need an equation. To find it, we start by calculating the rate of change:

[m = frac{1.81 - 1.45}{3 - 0} = frac{0.36}{3} = 0.12 ext{ million dollars per year.} onumber]

Since we already know the vertical intercept of the line, we can immediately write the equation:

[R(t) = 1.45 + 0.12t]

To predict the revenue in 2020, we evaluate our function at (t=8),

[R(8) = 1.45 + 0.12(8) = 2.41 onumber]

If the trend continues, our model predicts a revenue of $2.41 million in 2020.

To find when the population will reach $3 million, we can set (R(t)=3) and solve for (t).

[egin{align*} 3 &= 1.45 + 0.12t 1.55 &= 0.12t t &approx 12.917 end{align*} ]

Our model predicts the revenue will reach $3 million in a little under 13 years after 2012, or just before the year 2025.

In business, a very common application of functions is to model cost, revenue, and profit.

Definition: Word

When a company produces q items, the total cost is the cost of total cost of producing those items. The total cost includes both fixed costs, which are startup costs, like equipment and buildings, and variable costs, which are costs that depend on the number of items produced, like materials and labor.

In the most simple case, Total Cost = (Fixed Costs) + (Variable Costs) ∙ q

Revenue is the amount of money a company brings in from sales.

In the most simple case, Revenue = (Price per item) ∙ q

Profit is the amount of money brings in, after expenses.

Profit = Revenue – Costs

We often talk about the break-even point. This is the level of production where Revenue equals Cost, or equivalently where Profit is zero. This is typically the minimum level of sales necessary for the company to make a profit.

Example (PageIndex{4})

A tech startup is looking at developing and launching a new mobile app. Initial development of the app will cost $300,000, and they estimate marketing and support for each user will cost $0.50. While the app will be free, they estimate they will be able to bring in $2 per user on average from in-app purchases. How many users will the company need to break even?

We start by modeling the cost, revenue, and profit. Let q = number of users.

The fixed (initial) costs are $300,000, and the variable (per-item) costs as $0.50 per user. We can write the total cost equation:

[TC(q) = 300,000 + 0.50q onumber]

The revenue will be $2 per user, so the revenue equation will be:

[R(q) = 2q onumber]

We could find the break-even point by setting the total cost equal to the revenue, which is equivalent to finding the intersection of the lines.

Alternatively, we could go ahead and find a profit equation first:

[P(q) = R(q) - TC(q) = 2q - left( {300,000 + 0.50q} ight) = 1.5q - 300,000 onumber]

The break even point can be found by setting the profit equal to zero:

[egin{align*} 0 &= 1.5q - 300,000 q &= 200,000 end{align*} ]

The company will have to acquire 200,000 users to break even.

Exercise (PageIndex{1})

A donut shop estimates their fixed daily expenses are $600. If each donut costs about $0.05 to make and sells for $0.60, how many donuts do they need to sell to break even?


Revenue: (R = 0.60q)

Break even when (R=C), at a quantity of about 1091 donuts per day.

In economics, there is a model for how prices are determined in a free market which states that supply and demand for a product is related to the price. The demand relationship shows the quantity of a certain product that consumers are willing to buy at a certain price. Typically the quantity demanded will decrease for an item if the price increases. The supply relationship shows the quantity of a product that suppliers are willing to produce at a certain sales price. Typically the supply demanded will increase if the price increases. Economic theory says that supply and demand will interact, and the intersection will be the equilibrium price, or market price, where the quantity supplied and demanded will be equal.

Supply and Demand

If (p) is the price of a product, then

(Q_d) is the quantity demanded;

(Q_s) is the quantity supplied.

The demand curve is a decreasing function, while the supply curve is an increasing function.

The intersection of the curves is the equilibrium price and quantity, also called the market price and quantity. This point is often notated as (p^*, Q^*).

In later chapters you will explore supply and demand curves that are non-linear, but in this chapter we will focus on linear supply and demand functions.

In most economic books, you will see the supply and demand curve written with price as the input and quantity as the output, like (Q_d = 140 - 2p). However, supply and demand graphs are drawn with price on the vertical axis and quantity on the horizontal. In an effort to avoid confusion, most of the time in this text we will instead write supply and demand curves with price as the output, to match its placement on the vertical axis.

Example (PageIndex{5})

At a price of $2.50 per gallon, there is a demand in a certain town for 42.5 thousand gallons of gas and a supply of 20 thousand gallons. At a price of $3.50, there is demand for 25.5 thousand gallons and a supply of 28 thousand gallons. Assuming supply and demand are linear, find the equilibrium price and quantity.


We start by finding a linear equation for both supply and demand. We will use price, (p) in dollars, as the output and quantity, (q) in thousands of gallons, as the input.

For supply, we have the points ((20, 2.50)) and ((28, 3.50)).

Finding the slope: (m = frac{3.50 - 2.50}{28 - 20} = frac{1}{8})

We know the equation will look like (p = frac{1}{8}q + b), so substituting in ((20, 2.50)),

[egin{align*} 2.5 &= frac{1}{8}(20) + b 2.5 &= 2.5 + b b &= 0 end{align*} ]

The supply equation is: (p = frac{1}{8}q)

For demand, we have the points ((42.5, 2.50)) and ((25.5, 3.50)). Using a similar approach, we can find the demand equation is: (p = -frac{1}{17}q + 5).

To find the equilibrium, we set the supply equal to the demand:

[egin{align*} frac{1}{8}q &= -frac{1}17q + 5 &quad ext{Multiplying through by 8(17) = 136 to clear the fractions} & 136left( frac{1}{8}q ight) &= 136left( -frac{1}{17}q + 5 ight)& 17q &= - 8q + 680& quad ext{Now we solve for } q & 25q &= 680& q &= 27.2& end{align*} ]

To find the equilibrium price, we can substitute that value back into either equation:

[p = frac{1}{8}(27.2) = 3.4 onumber]

The equilibrium quantity will be 27.2 thousand gallons of gas at a price of $3.40.

Exercise (PageIndex{1})

A company estimates that at a price of $140 there will demand for 4000 items, and for each $5 increase in price the demand will drop by 200 items. The supply curve is (p = frac{1}{20}q). Find the equilibrium price and quantity.


Demand: (p = - frac{1}{40}q + 240)

Supply = Demand when (q = 3200, p =$160)

Important Topics of this Section

The problem solving process

  1. Identify changing quantities, and then carefully and clearly define descriptive variables to represent those quantities. Often this will involve checking and tracking units, building a table or even finding a formula for the function being used to model the problem.
  2. When needed, find a formula for the function.
  3. Solve or evaluate using the formula you found for the desired quantities.
  4. Clearly convey your result using appropriate units, and answer in full sentences when appropriate.


A manufacturer produces 80 units of a product at $22000 and 125 units at a cost of $28750. Assuming the cost curve to be linear, find the equation of the line and then use it to estimate the cost of 95 units. 

Since the cost curve is linear, its equation will be

(Here y = Total cost, x = no. of units)

80 units at $22000 -----> 22000  =  80A + B -----(1)

125 units at $28750 -----> 28750  =  125A + B -----(2)

Solving (1) and (2), we get A  = 150 and B = 10000

So, the equation of the line is

To find the cost of 95 units, substitute x = 95 in (3).

The cost of 95 units is $24250

A trader has 100 units of a product. A sells some of the units at $6 per unit and the remaining units at $8 per units. He receives a total of $660 for all 100 units. Find the number units sold in each category. 

x -----> number of units sold at $6/unit

y -----> number of units sold at $8/unit

The number of tickets sold at $6 per unit is 70 and  the number of tickets sold at $8 per unit is 30.

The wages of 8 men and 6 boys amount to $33. If 4 men earn $4.50 more than 5 boys, determine the wages of each man and boy. 

Let x and y be the wages of each man and boy. 

The wages of each man and each boy are $3 and $1.50 respectively

Sum of incomes of A and B is $2640. If B's income is 20% more than A, find the income of A and B. 

Let x and y be the incomes of A and B respectively.

Given :਋'s income is 20% more than A

The incomes of A and B are $1200 and $1440

A trader gains one third of the cost price as profit on a product and one fourth of the cost price as profit on other product. Total profit earned on these two products is $43.  The sum of the cost prices of two products is $150. Find the cost price of each product. 

Let x and y be the cost prices of the two products.

Given : One third of the cost price as profit on a product

So, profit on the first product = (1/3)x = x/3.

Given : One fourth of the cost price as profit on the other product. 

So, profit on the second product = (1/4)y = y/4.

Total profit earned on these two products = $43.

The cost prices of two products are $66 and $84 

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1.6: Modeling with Linear Functions - Mathematics

Linear functions can be used as models in the biological sciences when a particular dependent quantity changes at a constant rate with respect to an independent variable. From a modeling perspective, the equation, y (x) = mx + b, can be interpreted as follows,

Constant What it represents in a linear model
the constant b the initial amount of the quantity being modeled when the independent variable is time
The slope (m) the constant rate of change of the quantity being modeled with respect to the independent variable

There are many linear models in the life sciences. We will introduce the basic concepts of linear modeling by considering linear growth models.

Linear Growth Model

Organisms generally grow in spurts that are dependent on both environment and genetics. Under controlled laboratory conditions, however, one can often observe a constant rate of growth. These periods of constant growth are often referred to as the linear portions of the growth curve. In this example, we will construct a linear equation to model the linear phase of growth for a hypothetical insect larvae.

Suppose the larvae of a particular insect species grow linearly in mass during the last instar, from t = 0 to t = 48 hours. Laboratory observations indicate that, on average, the larvae are initially 8 grams and grow to an average size of 12 grams over the 48 hour period. To find the equation that models the growth, we must find the constants m and b in the equation,

The variable y represents larval mass in grams (g) and t represents time in hours (h). We place the restriction 0 &le t &le 48 on t because the model is only valid from hour 0 to hour 48. At hour 48, growth will begin to slow as the insect prepares to undergo pupation. Therefore, our model is only valid for the linear portion of the growth curve.

Now that we have our model, we can find the initial mass of the larvae (represented by the constant b) based on the information provided. We are given that the larvae are initially 8 g, therefore b is equal to 8 and we write,

To find the slope of the line (m), we take two points that we know to lie on the line. We choose the two points as (t1, y1) = (0, 8) and (t2, y2) = (48, 12) based on the information given. Using the formula for slope, we have,

Therefore, we can write the equation that models larval insect growth as,

Now that we have dealt with the numbers and variables, we must also make sure our model has consistent units. The units on each side of the equation must match- the dependent variable on the left side of the equation has units of grams, then the units on the right side must also be grams. Checking the consistency of the units is called dimensional analysis.

There are two terms on the right hand side, mt and b. Both of these terms must have units of g. Naturally, the term b, which represents the initial mass of the larvae, has units of g and t, which represents time, has units of hours. What are the units on m? We can determine m&rsquos units as follows,

where · represents m&rsquos units. Using this equation, we see that · must have units of g/hour in order for hours to cancel as,

Thus, we interpret our slope using the units as:

&ldquoLarvae grow 1 /12 th of a gram per hour.&rdquo

Of course, our model is only an approximation to reality there will exist variation in mass among larvae. For example, one larva may grow to 12.1 g while another grows to only 11.9 g over the 48 hours. We simply know that on average a larva grows to 12 g over the 48 hour time period.

Dimensional analysis can also provide an internal check of your work. If at the end of your calculations you find that "seconds = grams", you know that something is seriously wrong with either your model, your calculations, or both!

Population Growth Rate

Linear models can also be applied to the growth of a population of organisms. For example, the logistic population growth model assumes that the population growth rate (r) decreases linearly with population size (N) through a process known as intraspecific competition. The population growth rate can be positive, negative, or zero. If r is greater than zero the population grows, while if r is less than zero the population declines. When r = 0 the population size remains constant.

We can write an equation for r as a function of N as follows,

where r0 > 0 is the population growth rate in the absence of intraspecific competition and
> 0 is the carrying capacity of the environment. We can rewrite r(N) in slope-intercept form as,

In this example, the dependent variable is the population growth rate, r , and the independent variable is the population size, N, since r0 and K are positives numbers. The slope of the line is given by

the negative sign indicates that r decreases with N. The r-intercept (y-intercept) is given by b = r0 , which means that when N = 0, there is no intraspecific competition. We can find the N-intercept (x-intercept) of r (N) by setting r (N) = 0 and solving for N,

Therefore, we find the N-intercept to be K. We set r (N) equal to zero as above because the r-coordinate (y-coordinate) of points lying on the N-axis (x-axis) must be zero. We can plot r(N) as follows,

This graph indicates that

In other words, the population grows when the population is below the carrying capacity, and declines when the population is above the carrying capacity. When the population size is at the carrying capacity ( N = K), there is no growth or decline (r = 0) .

In the next section we will examine some other applications of linear models.

1.6 Mathematical Treatment of Measurement Results

It is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties:

An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of

Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation among the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:

The time can then be computed as:

Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/m/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.”

These calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method ). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers. This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.

Conversion Factors and Dimensional Analysis

A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor . For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,

Several other commonly used conversion factors are given in Table 1.6.

Length Volume Mass
1 m = 1.0936 yd 1 L = 1.0567 qt 1 kg = 2.2046 lb
1 in. = 2.54 cm (exact) 1 qt = 0.94635 L 1 lb = 453.59 g
1 km = 0.62137 mi 1 ft 3 = 28.317 L 1 (avoirdupois) oz = 28.349 g
1 mi = 1609.3 m 1 tbsp = 14.787 mL 1 (troy) oz = 31.103 g

When a quantity (such as distance in inches) is multiplied by an appropriate unit conversion factor, the quantity is converted to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by:

Example 1.8

Using a Unit Conversion Factor


The unit conversion factor may be represented as:

The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.

Check Your Learning


Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to ensure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. As your study of chemistry continues, you will encounter many opportunities to apply this approach.

Example 1.9

Computing Quantities from Measurement Results and Known Mathematical Relations


Volume may be converted from quarts to milliliters via two steps:

    Step 1. Convert quarts to liters.

Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:

Check Your Learning


Example 1.10

Computing Quantities from Measurement Results and Known Mathematical Relations

(a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?

(b) If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?


and then convert volume from liters to gallons:

Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:

(b) Using the previously calculated volume in gallons, we find:

Check Your Learning

(a) What (average) fuel economy, in miles per gallon, did the Prius get during this trip?

(b) If gasoline costs $3.90 per gallon, what was the fuel cost for this trip?


Conversion of Temperature Units

We use the word temperature to refer to the hotness or coldness of a substance. One way we measure a change in temperature is to use the fact that most substances expand when their temperature increases and contract when their temperature decreases. The mercury or alcohol in a common glass thermometer changes its volume as the temperature changes, and the position of the trapped liquid along a printed scale may be used as a measure of temperature.

Temperature scales are defined relative to selected reference temperatures: Two of the most commonly used are the freezing and boiling temperatures of water at a specified atmospheric pressure. On the Celsius scale, 0 °C is defined as the freezing temperature of water and 100 °C as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the Fahrenheit scale, the freezing point of water is defined as 32 °F and the boiling temperature as 212 °F. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees).

Defining the Celsius and Fahrenheit temperature scales as described in the previous paragraph results in a slightly more complex relationship between temperature values on these two scales than for different units of measure for other properties. Most measurement units for a given property are directly proportional to one another (y = mx). Using familiar length units as one example:

where y = length in feet, x = length in inches, and the proportionality constant, m, is the conversion factor. The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one (y = mx + b). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, m, it also must take into account differences in the scales’ zero points (b).

The linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as x and the Fahrenheit temperature as y, the slope, m, is computed to be:

The y-intercept of the equation, b, is then calculated using either of the equivalent temperature pairs, (100 °C, 212 °F) or (0 °C, 32 °F), as:

The equation relating the temperature (T) scales is then:

An abbreviated form of this equation that omits the measurement units is:

Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:

As mentioned earlier in this chapter, the SI unit of temperature is the kelvin (K). Unlike the Celsius and Fahrenheit scales, the kelvin scale is an absolute temperature scale in which 0 (zero) K corresponds to the lowest temperature that can theoretically be achieved. Since the kelvin temperature scale is absolute, a degree symbol is not included in the unit abbreviation, K. The early 19th-century discovery of the relationship between a gas’s volume and temperature suggested that the volume of a gas would be zero at −273.15 °C. In 1848, British physicist William Thompson, who later adopted the title of Lord Kelvin, proposed an absolute temperature scale based on this concept (further treatment of this topic is provided in this text’s chapter on gases).

The freezing temperature of water on this scale is 273.15 K and its boiling temperature is 373.15 K. Notice the numerical difference in these two reference temperatures is 100, the same as for the Celsius scale, and so the linear relation between these two temperature scales will exhibit a slope of 1 K °C 1 K °C . Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be:

The 273.15 in these equations has been determined experimentally, so it is not exact. Figure 1.28 shows the relationship among the three temperature scales.


Which kind of function best models the set of data points (–1, 22), (0, 6), (1, –10), (2, –26), and (3, –42)? linear **** quadratic exponential none of the above 2. Which kind of function best models the set of data points

Pre Algebra

The data in the table illustrate a linear function. x| –3, 0, 3, 6 y| –6, –2, 2, 6 What is the slope of the linear function? Which graph represents the data?

12. Which function is a quadratic function? A. y=3x^2+x B. y=2x-1 C. y=3/x D. y= -|x| Which function rule represents the data in the table? x -3, -2, -1, 0, 1 y 1, -2, -5, -8, -11 A. y=-3x-8 B. y=1/3x -8 C. y=1/3x+8 D. y=3x+8


which kind of function best models the data in the table use differences or ratios x y 0, 1.3 1, 7.8 2, 46.8 3, 280.8 4, 1684.8 A) linear B) quadratic C) exponential D) none of the above


Myra uses an inverse variation function to model the data for the ordered pairs below. (2, 30), (3, 20), (4, 15), (5, 12), (6, 10) Which statement best explains whether an inverse variation function is the best model for the data?

The table shows the outputs y for different inputs x: Input (x) 3 7 11 15 Output (y) 4 6 8 10 Part A: Do the data in this table represent a function? Justify your answer. (3 points) Part B: Compare the data in the table with the

Jane is organizing a fundraiser to buy a ping-pong table for the community center. The table costs $500.00. Jane is asking contributors to pay for an equal share of the cost of the table. She already has five contributors lined

Math I need help ASAP!

1) Identify the function rule shown in the table. Function Table n - 3, 4, 5, 6 y - 2, 1, 0, -1 a. y = 2 + n b. y = 5n c. y = 5 - n d. not enough information ** 2) What is the values of the function y = -2x - 4 for x = 0,1,2 and


Which kind of function best models the data in the table? Use differences or ratios. x y 0 1.7 1 6.8 2 27.2 3 108.8 4 435.2 A.linear B.quadratic C.exponential D.none of the above I think it is C.

Algebra 1

Does the data in the table represent a direct or inverse variation? Write an equation that models the data in the table. X | 1 | 3 | 5 | 10 Y | 4 | 12| 20| 40 A) direct variation y=4x B) direct variation y=1/4x C) inverse


kind of function best models the data on the table used to differences or ratios 0 0.6 1 4.2 2 29.4 3 205.8 4 1440.6


The function f is continuous on the interval [3, 13] with selected values of x and f(x) given in the table below. Use the data in the table to approximate f '(3.5) x 3 4 7 10 13 f(x) 2 8 10 12 22

Linear Algebra: Gateway to Mathematics: Second Edition

Linear Algebra: Gateway to Mathematics uses linear algebra as a vehicle to introduce students to the inner workings of mathematics. The structures and techniques of mathematics in turn provide an accessible framework to illustrate the powerful and beautiful results about vector spaces and linear transformations.

The unifying concepts of linear algebra reveal the analogies among three primary examples: Euclidean spaces, function spaces, and collections of matrices. Students are gently introduced to abstractions of higher mathematics through discussions of the logical structure of proofs, the need to translate terminology into notation, and efficient ways to discover and present proofs. Application of linear algebra and concrete examples tie the abstract concepts to familiar objects from algebra, geometry, calculus, and everyday life.

Students will finish a course using this text with an understanding of the basic results of linear algebra and an appreciation of the beauty and utility of mathematics. They will also be fortified with a degree of mathematical maturity required for subsequent courses in abstract algebra, real analysis, and elementary topology. Students who have prior background in dealing with the mechanical operations of vectors and matrices will benefit from seeing this material placed in a more general context.

Linear Equations Worksheet With Answers Pdf

1 6 r 7 13 7r 2 13 4x 1 x 3 7x 3x 2 8x 8 4 8 x x 4x. Let and represent algebraic expressions.

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The diagram below is a rectangle.

Linear equations worksheet with answers pdf. You can customize the worksheets to include one step two step or multi step equations variable on both sides parenthesis and more. Test and worksheet generators for math teachers. 25 x y 6 26 x 3 27 x 5 28 8x 7y 17 29 x 2y 12 30 2x 5y 5 write the standard form of the equation of each line given the slope and y intercept.

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Understanding the Cost Function

Let’s do an analysis using the squared error cost function.

Remember a cost function maps event or values of one or more variables onto a real number. In this case, the event we are finding the cost of is the difference between estimated values, or the difference between the hypothesis and the real values — the actual data we are trying to fit a line to.

Let’s unwrap the mess of greek symbols above. On the far left, we have 1/2*m. m is the number of samples — in this case, we have three samples for X. Those are 1 , 2 and 3 . So 1/2*m is a constant. It turns out to be 1/6 , or 0.1667 .

This means the sum. In this case, the sum from i to m , or 1 to 3 . We repeat the calculation to the right of the sigma, that is:

The actual calculation is just the hypothesis value for h(x) , minus the actual value of y . Then you square whatever you get.

The final result will be a single number. We repeat this process for all the hypothesis, in this case best_fit_1 , best_fit_2 and best_fit_3 . Whichever has the lowest result, or the lowest “cost” is the best fit of the three hypothesis.

Let’s go ahead and see this in action to get a better intuition for what’s happening.

Slope of Linear Functions

The concept of slope is important in economics because it is used to measure the rate at which changes are taking place. Economists often look at how things change and about how one item changes in response to a change in another item.

It may show for example how demand changes when price changes or how consumption changes when income changes or how quickly sales are growing.

Slope measures the rate of change in the dependent variable as the independent variable changes. The greater the slope the steeper the line.

Consider the linear function:

b is the slope of the line. Slope means that a unit change in x, the independent variable will result in a change in y by the amount of b.

slope = change in y/change in x = rise/run

Slope shows both steepness and direction. With positive slope the line moves upward when going from left to right. With negative slope the line moves down when going from left to right.

If two linear functions have the same slope they are parallel.

Slopes of linear functions

The slope of a linear function is the same no matter where on the line it is measured. (This is not true for non-linear functions.)

An example of the use of slope in economics

Demand might be represented by a linear demand function such as

Q(d) represents the demand for a good

P represents the price of that good.

Economists might consider how sensitive demand is to a change in price.

This is a typical downward sloping demand curve which says that demand declines as price rises.

This is a special case of a horizontal demand curve which says at any price above P* demand drops to zero. An example might be a competitor's product which is considered just as good.

This is a special case of a vertical demand curve which says that regardless of the price quantity demanded is the same. An example might be medicine as long as the price does not exceed what the consumer can afford.

Supply might be represented by a linear supply function such as

Q(s) represents the supply for a good

P represents the price of that good.

Economists might consider how sensitive supply is to a change in price.

This is a typical upward sloping supply curve which says that supply rises as price rises.

An example of the use of slope in economics

The demand for a breakfast cereal can be represented by the following equation where p is the price per box in dollars:

This means that for every increase of $1 in the price per box, demand decreases by 1,500 boxes.

Calculating the slope of a linear function

Slope measures the rate of change in the dependent variable as the independent variable changes. Mathematicians and economists often use the Greek capital letter D or D as the symbol for change. Slope shows the change in y or the change on the vertical axis versus the change in x or the change on the horizontal axis. It can be measured as the ratio of any two values of y versus any two values of x.

Generalized Linear Models (GLMs) in R, Part 4: Options, Link Functions, and Interpretation

Last year I wrote several articles (GLM in R 1, GLM in R 2, GLM in R 3) that provided an introduction to Generalized Linear Models (GLMs) in R.

As a reminder, Generalized Linear Models are an extension of linear regression models that allow the dependent variable to be non-normal.

In our example for this week we fit a GLM to a set of education-related data.

Let’s read in a data set from an experiment consisting of numeracy test scores (numeracy), scores on an anxiety test (anxiety), and a binary outcome variable (success) that records whether or not the students eventually succeeded in gaining admission to a prestigious university through an admissions test.

We will use the glm() command to run a logistic regression, regressing success on the numeracy and anxiety scores.

The variable ‘success’ is a binary variable that takes the value 1 for individuals who succeeded in gaining admission, and the value 0 for those who did not. Let’s look at the mean values of numeracy and anxiety.

We begin by fitting a model that includes interactions through the asterisk formula operator. The most commonly used link for binary outcome variables is the logit link, though other links can be used.

glm() is the function that tells R to run a generalized linear model.

Inside the parentheses we give R important information about the model. To the left of the

is the dependent variable: success. It must be coded 0 & 1 for glm to read it as binary.

, we list the two predictor variables. The * indicates that not only do we want each main effect, but we also want an interaction term between numeracy and anxiety.

And finally, after the comma, we specify that the distribution is binomial. The default link function in glm for a binomial outcome variable is the logit. More on that below.

We can access the model output using summary() .

The estimates (coefficients of the predictors – numeracy and anxiety) are now in logits. The coefficient of numeracy is: 1.94556, so that a one unit change in numeracy produces approximately a 1.95 unit change in the log odds (i.e. a 1.95 unit change in the logit).

From the signs of the two predictors, we see that numeracy influences admission positively, but anxiety influences survival negatively.

We can’t tell much more than that as most of us can’t think in terms of logits. Instead we can convert these logits to odds ratios.

We do this by exponentiating each coefficient. (This means raise the value e –approximately 2.72–to the power of the coefficient. e^b).

So, the odds ratio for numeracy is:

However, in this version of the model the estimates are non-significant, and we have a non-significant interaction. Model1 produces the following relationship between the logit (log odds) and the two predictors:

logit(p) = 0.88 + 1.95* numeracy – 0.45 * anxiety – 1.0* interaction term

The output produced by glm() includes several additional quantities that require discussion.

We see a z value for each estimate. The z value is the Wald statistic that tests the hypothesis that the estimate is zero. The null hypothesis is that the estimate has a normal distribution with mean zero and standard deviation of 1. The quoted p-value, P(>|z|), gives the tail area in a two-tailed test.

For our example, we have a Null Deviance of about 68.03 on 49 degrees of freedom. This value indicates poor fit (a significant difference between fitted values and observed values). Including the independent variables (numeracy and anxiety) decreased the deviance by nearly 40 points on 3 degrees of freedom. The Residual Deviance is 28.2 on 46 degrees of freedom (i.e. a loss of

About the Author: David Lillis has taught R to many researchers and statisticians. His company, Sigma Statistics and Research Limited, provides both on-line instruction and face-to-face workshops on R, and coding services in R. David holds a doctorate in applied statistics.

Watch the video: PreCalculus - Section - Modeling with Linear Functions (October 2021).