Answer
a. $2078\ cal/day$, underestimates by $22$ calories.
b. $2662\ cal/day$, underestimates by $38$ calories.
c. $\frac{W}{M}=\frac{ -33x^2 + 263x + 515}{ -60x^2 +499x + 295}$
Work Step by Step
a. We are given the model equation
$W = -66x^2 + 526x + 1030$
Ages 19 to 30 belong to group 4, so we have $x=4$ and
$W = -66(4)^2 + 526(4) + 1030=2078\ cal/day$
Reading from the bar graph for women of this group, we get the number $2100$; thus the model underestimates the real value by $22$ calories.
b. Given the model equation
$M = -120x^2 + 998x + 590$
ages 19 to 30 belong to group 4. Thus We have
$x=4$ and $M = -120(4)^2 + 998(4) + 590=2662\ cal/day$
Reading from the bar graph for men of this group, we get a number $2700$; thus the model underestimates the real value by $38$ calories.
c. Taking the ratio of the two equations, we have
$\frac{W}{M}=\frac{ -66x^2 + 526x + 1030}{ -120x^2 + 998x + 590}$
Simplify by cancelling common factors; we have
$\frac{W}{M}=\frac{ -33x^2 + 263x + 515}{ -60x^2 +499x + 295}$
