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| Section | Weight | Objectives |
|---|---|---|
| Systems of Equations and Inequalities | 15% | - Graphical solutions - Solving by substitution and elimination - Applications of systems |
| Exponents, Radicals, and Quadratic Relationships | 15% | - Properties of exponents - Simplifying radical expressions - Basic quadratic graphs - Solving quadratic equations |
| Algebraic Expressions and Operations | 20% | - Variables, constants, and coefficients - Simplifying and evaluating expressions - Operations with polynomials - Order of operations |
| Linear Equations and Inequalities | 25% | - Real-world applications - Solving and graphing inequalities - Solving single-variable equations |
| Graphing and Functions | 25% | - Linear functions and their graphs - Function notation and evaluation - Slope and equations of lines - Coordinate plane and plotting points |
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NEW QUESTION # 82
A vehicle is traveling away from a town at a fixed rate. After 1 hours, the vehicle is 200 miles from the town.
After 4 hours, the vehicle is 395 miles from the town.
Which function represents the distance, d, between the vehicle and the town after thours?
Answer: C
Explanation:
Because the vehicle is traveling away from the town at a fixed rate, the distance can be modeled by a linear function:
d(t)=mt+b
where:
m= " rate of change "
and
b= " initial distance from the town "
We are given two points:
(1#200)
and
(4#395)
These mean:
After 1hour, the vehicle is 200miles away.
After 4hours, the vehicle is 395miles away.
First, find the rate of change:
m=(395-200)/(4-1)
m=195/3
m=65
So the vehicle is moving away from the town at a rate of:
65 " miles per hour "
Now the function has the form:
d(t)=65t+b
Use the point (1#200)to find b:
200=65(1)+b
200=65+b
b=135
Therefore, the function is:
d(t)=65t+135
Check using t=4:
d(4)=65(4)+135
d(4)=260+135
d(4)=395
NEW QUESTION # 83
The data in the scatterplot represents the number of people working at a research station over time. The graphed regression function has an r
2
value of 0.96. The predicted number of people working at the research station after 18.1 years is 541.
Is this prediction appropriate?
Answer: A
Explanation:
The value r2=0.96 indicates a strong regression fit because it is very close to 1. A strong fit supports prediction better than a weak or moderate fit. However, prediction also depends on whether the x-value is within a reasonable extrapolation range. The prediction is made at x=18.1, which is beyond the maximum observed x-value in the scatterplot. Even though the model fits the data well, the prediction is too far beyond the observed data range.
For a strong model, extrapolation should not go more than about 50% of the range beyond the maximum value. Since x=18.1 exceeds that limit, the prediction is not appropriate. Therefore, the correct answer is B.
NEW QUESTION # 84
The graphed function v(t) represents the number of vehicles, v, stopped at a toll booth t hours after 6:00 a.m.
The coordinates of points A and B are (1,49.4) and (4,45.8), respectively.
What is the average rate of change of the number of vehicles from point A to point B?
Answer: D
Explanation:
The average rate of change measures how much the output changes per one unit of input. Here, the input is time in hours after 6:00 a.m., and the output is the number of vehicles stopped at the toll booth. The two given points are A=(1,49.4) and B=(4,45.8). Use the average rate of change formula: (y
2
#y
1
)/(x
2
#x
1
). Substituting the coordinates gives (45.8#49.4)/(4#1). The numerator is #3.6, and the denominator is 3, so the average rate of change is #1.2. The negative sign means the number of vehicles decreased over the interval.
NEW QUESTION # 85
The graphed function n(t)represents the number of people, n, at a park thours after 8:00 a.m. The plotted points Aand Bhave coordinates (2#110)and (6#182).
Which statement gives the correct interpretation of the average rate of change of the number of people over the interval from point Ato point B?
Answer: D
Explanation:
The average rate of change measures how much the output changes per one unit of input.
Here:
t= " hours after 8:00 a.m. "
and
n(t)= " number of people at the park "
The two given points are:
A=(2,110)
and
B=(6,182)
This means:
At t=2hours after 8:00 a.m., there were 110people at the park.
At t=6hours after 8:00 a.m., there were 182people at the park.
Now calculate the average rate of change:
" Average rate of change " =(182-110)/(6-2)
=72/4
=18
Because the value is positive, the number of people is increasing.
So the correct interpretation is:
" The number of people at the park increases at an average rate of 18 people per hour. "
NEW QUESTION # 86
A new company just launched and is using the function M(t)to predict its market share after tyears. The graph of M(t)is shown.
When should the company expect to have a market share of 40%?
Answer: C
Explanation:
The function M(t)represents the company's market share after tyears.
The horizontal axis represents:
t= " time in years "
The vertical axis represents:
M(t)= " market share percentage "
We need to find when the market share reaches:
40%
From the graph, locate 40on the vertical axis. Then move horizontally to the blue curve and read the corresponding value on the horizontal axis.
The curve reaches 40%at approximately:
t=7.4
So the company should expect to have a market share of 40%after about:
7.4 " years "
Therefore, the correct answer is:
# ( " C " )
NEW QUESTION # 87
......
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