Question

question 3 (1 point)
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sasha bought a car for $32,875. seven years later, the car was worth $22,095. if the depreciation was linear (straight - line depreciation method), by how much did the value of the car go down each year?
what would the car have been worth after three years?
decreases $1540 per year; worth $28,255 after three years
decreases $3156.43 per year; worth $23,405.71 after three years
decreases $4696.43 per year; worth $14,089.29 after three years
decreases $1540 per year; worth $4,620 after three years
decreases $4696.43 per year; worth $18,785.71 after three years
question 4 (1 point)
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a school has 1560 students. four years later, the same school has 1640 students. if there was linear growth, the slope of this line would (numerically) be 20. explain what this means.
the number of students at the school increased by 20 each year.
the number of students at the school decreased by 20 each year.
every 20 years, there will be one more student at the school.
the number of students at the school increased by 20 every four years.
question 5 (1 point)
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suppose you put $500 in the bank, and it will earn 4% interest compounded each year. choose the answer that shows how much you will have in the bank after 1 year, 2 years, and 3 years (answers are all rounded to the nearest cent).
$700, $980, $1372
$520, $540, $560
$520, $540.80, $562.43

Explanation:

Step1: Calculate annual depreciation for Question 3

The formula for annual depreciation in straight - line method is \(D=\frac{P - V}{n}\), where \(P\) is the initial price (\(P = 32875\)), \(V\) is the final value (\(V = 22095\)), and \(n\) is the number of years (\(n = 7\)).

$$D=\frac{32875 - 22095}{7}=\frac{10780}{7}=1540$$

Step2: Calculate the value after 3 years for Question 3

The formula for the value \(A\) of the car after \(t\) years is \(A=P - D\times t\). Here, \(P = 32875\), \(D = 1540\), and \(t = 3\)

$$A=32875-1540\times3=32875 - 4620=28255$$

Step3: Analyze the slope for Question 4

The slope formula for a linear relationship \(y=mx + b\) (where \(y\) is the number of students, \(x\) is the number of years) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). A positive slope means growth. If \(m = 20\), it means the number of students increases by 20 each year.

Step4: Calculate compound - interest for Question 5

The compound - interest formula is \(A=P(1 + r)^t\), where \(P = 500\), \(r=0.04\)

• For \(t = 1\): \(A=500(1 + 0.04)^1=500\times1.04 = 520\)
• For \(t = 2\): \(A=500(1 + 0.04)^2=500\times(1.04)^2=500\times1.0816 = 540.80\)
• For \(t = 3\): \(A=500(1 + 0.04)^3=500\times(1.04)^3=500\times1.124864\approx562.43\)

Answer:

Question 3: Decreases \(\$1540\) per year; Worth \(\$28,255\) after three years
Question 4: The number of students at the school increased by 20 each year
Question 5: \(\$520\), \(\$540.80\), \(\$562.43\)