Question

The percentage of obese children aged 12-19 years in the United States is approximately P(t) = 0.04t + 4.6 if 0 ≤ t < 10 −0.01005t2 + 0.945t − 3.4 if 10 ≤ t ≤ 30 where t is measured in years, with t = 0 corresponding to the beginning of 1970. What was the percentage of obese children aged 12-19 at the beginning of 1975? At the beginning of 1985? At the beginning of 1990?† (Round your answers to two decimal places.) 1975 % 1985 % 1990 %

Answer 1

Answer:

1975------ 4.8%

1990 ------ 11.48%

1985 ------ 8.51%

Step-by-step explanation:

Given

$P(t) = 0.04t + 4.6$ if 0 ≤ t < 10

$P(t) = -0.01005t^2 + 0.945t - 3.4$ if 10 ≤ t ≤ 30

Where

t:=0 implies year = 1970

Solving (a): Beginning of 1975.

First, we calculate the value of t

$t = 1975 - 1970$

$t = 5$

This falls in the range: 0 ≤ t < 10

So:

$P(t) = 0.04t + 4.6$

$P(5) = 0.04 * 5 + 4.6$

$P(5) = 4.80$

Solving (b): Beginning of 1990.

$t = 1990 - 1970$

$t = 20$

This falls in the range: 10 ≤ t ≤ 30

So:

$P(t) = -0.01005t^2 + 0.945t - 3.4$

$P(20) = -0.01005*20^2 + 0.945*20 - 3.4$

$P(20) = 11.48$

Solving (c): Beginning of 1985

$t = 1985 - 1970$

$t = 15$

This falls in the range: 10 ≤ t ≤ 30

So:

$P(t) = -0.01005t^2 + 0.945t - 3.4$

$P(15) = -0.01005*15^2 + 0.945*15 - 3.4$

$P(15) = 8.51375$

$P(15) = 8.51$