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Math  /  Data & Statistics

Question5. Joey enlarged a 3 -inch by 5 -inch photograph on a copy machine. He enlarged it four times. The table below shows the area of the photograph after each enlargement. \begin{tabular}{|l|c|c|c|c|c|} \hline Enlargement & 0 & 1 & 2 & 3 & 4×24 \times 2 \\ \hline Area (square inches) & 15y15 y & 18.8 & 23.4 & 29.3 & 36.6y36.6 y \\ \hline \end{tabular}
What is the average rate of change of the area from the original photograph to the fourth enlargement, to the nearest tenth? 4.34.55.46.036.6−154−0=21.64\begin{array}{ll} 4.3 \\ 4.5 \\ 5.4 \\ 6.0 \end{array} \quad \frac{36.6-15}{4-0}=\frac{21.6}{4}

Studdy Solution

STEP 1

1. The original area of the photograph is 15y 15y square inches.
2. The area after the fourth enlargement is 36.6y 36.6y square inches.
3. The average rate of change is calculated as the change in area divided by the change in the number of enlargements.

STEP 2

1. Identify the initial and final areas.
2. Calculate the change in area.
3. Calculate the change in the number of enlargements.
4. Compute the average rate of change.

STEP 3

Identify the initial area (A0 A_0 ) and the final area (A4 A_4 ):
A0=15y A_0 = 15y A4=36.6y A_4 = 36.6y

STEP 4

Calculate the change in area (ΔA \Delta A ):
ΔA=A4−A0=36.6y−15y=21.6y \Delta A = A_4 - A_0 = 36.6y - 15y = 21.6y

STEP 5

Calculate the change in the number of enlargements (Δn \Delta n ):
Δn=4−0=4 \Delta n = 4 - 0 = 4

STEP 6

Compute the average rate of change (R R ):
R=ΔAΔn=21.6y4=5.4y R = \frac{\Delta A}{\Delta n} = \frac{21.6y}{4} = 5.4y
Since y y is consistent across all calculations, it cancels out, and the average rate of change is:
5.4 \boxed{5.4}

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