The relationship between the percentage of frozen citrus crop, and the cost of box of oranges is an illustration of a linear function.

The linear equation of the function is: $g(P) = 22.9P+7$ .
The inverse function is: $g^{-1}(c) = \frac{1}{22.9}(c - 7)$ .
A practical domain is from 0% to 100%
A practical range is from 7 to 29.9

A. Input quantity

The input quantity is the percentage of frozen citrus crop

B. Output quantity

The output quantity is the cost of box of oranges

C. The linear function

We have:

$(P_1,c_1) = (20\%,11.58)\\(P_2,c_2) = (80\%,25.32)$

Calculate the slope of the function

$m = \frac{c_2 - c_1}{P_2 - P_1}$

$m = \frac{25.32 - 11.58}{80\%-20\%}$

$m = \frac{13.74}{60\%}$

$m = 22.9$

The linear equation is calculated as follows:

$c -c_1 = m(P-P_1)$

$c -11.58= 22.9(P-20\%)$

$c-11.58 = 22.9P-4.58$

D. Rewrite as y = mx + b

We have:

$c-11.58 = 22.9P-4.58$

Collect like terms

$c = 22.9P - 4.58 + 11.58$

$c = 22.9P+7$

The function is:

$g(P) = 22.9P+7$

E. A practical domain

The domain is the possible values of P. Because P is a percentage, its possible values are 0% to 100%.

The domain of the function is: $[0\%,100\%]$

F. A practical range

When P = 0%

$c = 22.9 \times 0\% + 7 = 7$

When P = 100%

$c = 22.9 \times 100\% + 7 = 29.9$

Hence, the range of the function is: $[7,29.9]$

G. The meaning of $g^{-1}(12)$

The inverse function of g(P) is $g^{-1}(P)$

So:

$g^{-1}(12)$ is the percentage of frozen citrus crop, when the cost is $12.

H. The inverse formula

We have:

$c = 22.9P+7$

Subtract 7 from both sides

$c - 7 = 22.9P$

Make P the subject

$P = \frac{1}{22.9}(c - 7)$

So, the inverse formula is:

$g^{-1}(c) = \frac{1}{22.9}(c - 7)$

Substitute 12 for c

$g^{-1}(12) = \frac{1}{22.9}(12 - 7)$

$g^{-1}(12) = \frac{1}{22.9} \times 5$

$g^{-1}(12) = 22\%$

Read more about linear equations at:

brainly.com/question/19770987

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3 years ago

Algebra 1 CR-8

antiseptic1488 [7]

$\mathrm{Domain\:of\:}\:x^2+2x-15\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

$\mathrm{Range\:of\:}x^2+2x-15:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:-16\:\\ \:\mathrm{Interval\:Notation:}&\:[-16,\:\infty \:)\end{bmatrix}$

$\mathrm{Axis\:interception\:points\:of}\:x^2+2x-15:\quad \mathrm{X\:Intercepts}:\:\left(3,\:0\right),\:\left(-5,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:-15\right)$