Question

1. the table shows the tuition costs for a private school x years after it opens.

year, x: 0, 1, 2, 3
tuition (dollars), y: 36208, 37620, 39088, 40594
a. verify that the data show a linear relationship. then write an equation of a line of fit.
b. interpret the slope and y - intercept in this situation.
c. predict the cost of tuition after 5 years.

1. the area a of a trapezoid is represented by the formula a=\frac{1}{2}h(b_1 + b_2). which equation represents the height h of a trapezoid?

(a) h=\frac{a}{2(b_1 + b_2)}
(b) h=\frac{2a}{b_1 + b_2}
(c) h = 2a-(b_1 + b_2)
(d) h = 2a(b_1 + b_2)

1. in a trivia game, each correct answer is worth 10 points in the first round, 15 points in the second round, and 25 points in the third round. you score 370 points by answering 24 questions correctly. you score 55 more points in the third round than you do in the first round. how many questions do you answer correctly in each round?

Explanation:

8.

Step1: Calculate the rate of change

To verify linearity, find the differences in $y$ - values for a constant change in $x$ - values.
When $x$ changes from $0$ to $1$ ($\Delta x = 1$), $y$ changes from $36208$ to $37620$, so $\Delta y=37620 - 36208 = 1412$.
When $x$ changes from $1$ to $2$ ($\Delta x = 1$), $y$ changes from $37620$ to $39088$, so $\Delta y=39088 - 37620 = 1468$.
When $x$ changes from $2$ to $3$ ($\Delta x = 1$), $y$ changes from $39088$ to $40594$, so $\Delta y=40594 - 39088 = 1506$.
The differences are not exactly the same, but we can still find a line - of - fit using the least - squares method or the slope - intercept form.
The slope - intercept form of a line is $y=mx + b$, where $m$ is the slope and $b$ is the $y$ - intercept. Using the first two points $(0,36208)$ and $(1,37620)$:
The slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{37620 - 36208}{1-0}=1412$. The $y$ - intercept $b$ is the value of $y$ when $x = 0$, so $b = 36208$. The equation of the line of fit is $y=1412x + 36208$.

Step2: Interpret the slope and y - intercept

The slope $m = 1412$ means that the tuition cost increases by approximately $\$1412$ per year. The $y$ - intercept $b = 36208$ means that the initial tuition cost (when the school opens, $x = 0$) is $\$36208$.

Step3: Predict the tuition cost after 5 years

Substitute $x = 5$ into the equation $y=1412x + 36208$.
$y=1412\times5+36208=7060 + 36208=43268$.

We start with the formula for the area of a trapezoid $A=\frac{1}{2}h(b_1 + b_2)$.
First, multiply both sides of the equation by $2$ to get $2A=h(b_1 + b_2)$.
Then, divide both sides of the equation by $(b_1 + b_2)$ to solve for $h$. So $h=\frac{2A}{b_1 + b_2}$.

Let $x$ be the number of correct answers in the first round, $y$ be the number of correct answers in the second round, and $z$ be the number of correct answers in the third round.
We have the following system of equations:
Equation 1: Total number of questions
$x + y+z=24$
Equation 2: Total number of points
$10x + 15y+25z=370$
Equation 3: Relationship between first - and third - round points
$25z=10x + 55$ or $10x-25z=- 55$

From the first equation, $y = 24-(x + z)$.
Substitute $y = 24-(x + z)$ into the second equation:
$10x+15(24-(x + z))+25z=370$
$10x+360-15x-15z + 25z=370$
$-5x + 10z=10$ or $-x + 2z=2$, which can be rewritten as $x = 2z-2$.
Substitute $x = 2z-2$ into $10x-25z=-55$:
$10(2z-2)-25z=-55$
$20z-20-25z=-55$
$-5z=-35$
$z = 7$.
If $z = 7$, then $x=2z - 2=2\times7-2 = 12$.
Since $x + y+z=24$, then $12+y + 7=24$, so $y=24-(12 + 7)=5$.

Answer:

a. The data is approximately linear. The equation of the line of fit is $y = 1412x+36208$.
b. The slope of $1412$ means the tuition increases by about $\$1412$ per year, and the $y$ - intercept of $36208$ means the initial tuition is $\$36208$.
c. The predicted cost of tuition after 5 years is $\$43268$.

9.