Question

1. constructed response - 4 points - show all work for each question

alaine bought a car for $19,000. the value of the car depreciates at a rate of 9.4% each year.
a. write a function, v(t) that represents the value of the car t years after alaine bought it.
b. determine to the nearest cent how much the car was worth after 4 years
c. determine to the nearest cent how much the car was worth after 10 years
d. how much more was the car worth in 4 years than in 10 years?

1. selected response - 1 point

the model below represents the number of students at north county high school from 2000-2015 where s(t) is the number of students and t is the number of years since 2000.
$s(t)=1962(1.012)^t$
based on this model, which of the following statements are false?
a. the student population in 2000 was 1,962 students.
b. the student population grew by 12% each year during this 15-year period.
c. in 2015, the student population was approximately 2346 students.
d. the student population was increasing over this time period.

Explanation:

Question 6

Step1: Define depreciation function

For exponential depreciation, the formula is $V(t) = P(1-r)^t$, where $P=\$19000$, $r=0.094$.
$V(t) = 19000(1-0.094)^t = 19000(0.906)^t$

Step2: Calculate value after 4 years

Substitute $t=4$ into $V(t)$.
$V(4) = 19000(0.906)^4$
First compute $(0.906)^4 \approx 0.906\times0.906\times0.906\times0.906 \approx 0.6725$
$V(4) \approx 19000\times0.6725 = 12777.50$

Step3: Calculate value after 10 years

Substitute $t=10$ into $V(t)$.
$V(10) = 19000(0.906)^{10}$
First compute $(0.906)^{10} \approx 0.3827$
$V(10) \approx 19000\times0.3827 = 7271.30$

Step4: Find difference in values

Subtract $V(10)$ from $V(4)$.
$12777.50 - 7271.30 = 5506.20$

• For option a: When $t=0$ (2000), $S(0)=1962(1.012)^0=1962$, so this is true.
• For option b: The growth factor is $1.012$, so annual growth rate is $1.012-1=0.012=1.2\%$, not 12%, so this is false.
• For option c: 2015 is $t=15$, $S(15)=1962(1.012)^{15}\approx1962\times1.195\approx2346$, so this is true.
• For option d: The growth factor $1.012>1$, so the population is increasing, this is true.

Answer:

a. $V(t) = 19000(0.906)^t$
b. $\$12777.50$
c. $\$7271.30$
d. $\$5506.20$

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Question 7