In the graphic below LAE = KEA by:

A devastating freeze in California's Central Valley in January 2007 wiped out approximately 75% of the state's citrus crop. It t

solniwko [45]

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

6 0

3 years ago

In the adjoining figure O is the centre of the circle the tangent to the circle of radius 6 cm from the exterior point P of the

slamgirl [31]

Answer:

Let P be the external point. O be the origin. join O and P get OP and nearest point on the circle from P be A.

Let Q be the point onthe circle in which, tangent make 90° with radius at Q.

PQ = 8 and OQ = 6

we get a right angled triangle PQO right angled at Q.

so, OP^2 = OQ^2 + PQ^2= 8^2 + 6^2 = 64 + 36 =1==

therefore OP =10cm

we need nearest point from P, which is PA

PA = OP - OA= 10 -6=4cm

7 0

3 years ago

How do I do this problem? Evaluate each expression (-4)-(-6)-(2-1)

natita [175]

Answer:

1

Step-by-step explanation:

(-4) - (-6) - (2-1)

-4 + 6 -2 +1

1

6 0

3 years ago

Are they congruent? How do you know?

pogonyaev

Answer:

Yes, if you translate either plane, they will end up in the exact location.

5 0

3 years ago

The table gives the probabilities of which type of sandwich is likely to be the top-selling item each month in restaurants acros

spin [16.1K]

2,3,4,1, the probability is in order from greatest to least. If they asked for least to greatest then, 1,4,3,2. So use the least to greatest one.

7 0

3 years ago

Read 2 more answers