Graphing Data

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varies directly with $x^{2}$ . If , what is if ?

Explanation

1. Since varies directly with $x^{2}$ :

$y=Kx^{2}$ with K being some constant.

2. Solve for K using the x and y values given:

$75=K(5^{2})$

3. Use the equation you solved for to find the value of y:

$y=3x^{2}$

$y=3(10^{2})$

varies directly with $x^{2}$ . If , what is if ?

Explanation

1. Since varies directly with $x^{2}$ :

$y=Kx^{2}$ with K being some constant.

2. Solve for K using the x and y values given:

$75=K(5^{2})$

3. Use the equation you solved for to find the value of y:

$y=3x^{2}$

$y=3(10^{2})$

varies directly with $x^{2}$ . If , what is if ?

Explanation

1. Since varies directly with $x^{2}$ :

$y=Kx^{2}$ with K being some constant.

2. Solve for K using the x and y values given:

$75=K(5^{2})$

3. Use the equation you solved for to find the value of y:

$y=3x^{2}$

$y=3(10^{2})$

varies inversely with $x^{3}$ . If , then what is equal to when ?

Explanation

1. Since varies indirectly with $x^{3}$ :

$y=\frac{K}{x^{3}}$

2. Use the given x and y values to determine the value of K:

$2=\frac{K}{2^{3}}$

$2=\frac{K}{8}$

3. Using the equation along with the value of K, find the value of y when x=5:

$y=\frac{16}{x^{3}}$

$y=\frac{16}{5^{3}}$

The output of a factory in units per day versus the number of employees working is plotted on the graph below, with the following data points collected:

(Workers, Units of output per day):

$\left(100,1300\right)$

$\left(75,1050\right)$

$\left(200,2300\right)$

$\left(225,2550\right)$

Assuming a linear relationship, interpolate to find how many units will be made per day if workers are present.

Linear_interp

$1400\textup{ units}$

$1335\textup{ units}$

$1375\textup{ units}$

$1380\textup{ units}$

$1370\textup{ units}$

Explanation

We want to do a linear interpolation since the relationship between workers and units can be assumed to be linear. This means there is a constant slope between the points, so the slope between two known points will be equal to the slope between the point we are trying to find and some known point. This is expressed in the relation:

$\frac{y-y_0}{x-x_0}=\frac{y_1-y_0}{x_1-x_0}$ ,

where and are the points we want to find and and are known. We choose the known points to be those that are just to the left and right of the point we are trying to find,

and .

Plugging these into our interpolation formula and knowing , we can find , the units output per day.

$\frac{y-1300}{110-100}=\frac{2300-1300}{200-100}$ .

Simplifying and rearranging to solve for :

So there are units produced when the number of workers is .

The output of a factory in units per day versus the number of employees working is plotted on the graph below, with the following data points collected:

(Workers, Units of output per day):

$\left(100,1300\right)$

$\left(75,1050\right)$

$\left(200,2300\right)$

$\left(225,2550\right)$

Assuming a linear relationship, interpolate to find how many units will be made per day if workers are present.

Linear_interp

$1400\textup{ units}$

$1335\textup{ units}$

$1375\textup{ units}$

$1380\textup{ units}$

$1370\textup{ units}$

Explanation

$\frac{y-y_0}{x-x_0}=\frac{y_1-y_0}{x_1-x_0}$ ,

where and are the points we want to find and and are known. We choose the known points to be those that are just to the left and right of the point we are trying to find,

and .

Plugging these into our interpolation formula and knowing , we can find , the units output per day.

$\frac{y-1300}{110-100}=\frac{2300-1300}{200-100}$ .

Simplifying and rearranging to solve for :

So there are units produced when the number of workers is .

The output of a factory in units per day versus the number of employees working is plotted on the graph below, with the following data points collected:

(Workers, Units of output per day):

$\left(100,1300\right)$

$\left(75,1050\right)$

$\left(200,2300\right)$

$\left(225,2550\right)$

Assuming a linear relationship, interpolate to find how many units will be made per day if workers are present.

Linear_interp

$1400\textup{ units}$

$1335\textup{ units}$

$1375\textup{ units}$

$1380\textup{ units}$

$1370\textup{ units}$

Explanation

$\frac{y-y_0}{x-x_0}=\frac{y_1-y_0}{x_1-x_0}$ ,

where and are the points we want to find and and are known. We choose the known points to be those that are just to the left and right of the point we are trying to find,

and .

Plugging these into our interpolation formula and knowing , we can find , the units output per day.

$\frac{y-1300}{110-100}=\frac{2300-1300}{200-100}$ .

Simplifying and rearranging to solve for :

So there are units produced when the number of workers is .

A student scores a on the Scholastic Assessment Test (SAT). A college admissions committee does not know how the exam is scored; however, they do know the scores of the exam form a normal distribution pattern. They also know the mean score and standard deviation of the population of students that took the test.

$\textup{Mean}=\bar{x}=495$

$\textup{Standard deviation}=S_{x}=116$

Using this information, determine whether or not the student scored well on the SAT.

The student scored very well: above two standard deviations of the mean.

The student scored very poorly: below two standard deviations of the mean.

The student scored well: above a single standard deviation of the mean.

The student scored poorly: below a single standard deviation of the mean.

The student's score was average: similar to the mean.

Explanation

In order to solve this problem let's consider probabilities and the normal—bell curve—distribution. Given that all events are equally likely, probability is calculated using the following formula:

$\textup{Probability}=\frac{\textup{number of favorable outcomes}}{\textup{total number of outcomes}}$

When probabilities of a given population are calculated for particular events, they can be graphed in a frequency chart or histogram. If they form a standard distribution, then the graph will form to the following shape:

Normaldistribution

This shape is known as a bell curve. In this curve, the mean is known as the arithmetic average and is represented as the peak. The mean alters the position of the graph. If the mean increases or decreases, then the graph shifts to the right or to the left respectively. The mean is denoted as follows:

$\textup{Mean}=\bar{x}$

On the other hand, the standard deviation is a calculation that indicates the average amount that each value deviates from the mean. When the standard deviation is changed then the shape of the graph is altered. When the standard deviation is decreased, the graph is taller and thinner. Likewise, when the standard deviation is increased, the graph becomes shorter and wider. It is important to note that 99.7 percent of all the values in a normal population exist between three standard deviations above and below the mean. It is denoted using the following annotation:

$\textup{Standard deviation}=S_{x}$

Now that we have discussed the components of the bell curve, let's consider the scenario presented in the question.

We know that the distribution of test scores follows a normal curve. We also know the following values:

$\bar{x}=495$

$S_{x}=116$

We should first plot the data on a graph that follows the shape of a bell shaped curve with three standard deviations.

Satnorm

We know that the student had the following score:

$\textup{Student's score}=749$

Let's calculate two standard deviations above the mean.

$(2\times S_{x})+\bar{x}=(2\times116)+495=727$

The student scored very well: above two standard deviations from the mean. Notice that at this point on the graph, the tail of the curve is closer to the horizontal or x-axis. This means that fewer students scored this high on the exam. In other words, the student performed very well.

varies inversely with $x^{3}$ . If , then what is equal to when ?