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2 years ago

HELP PLEASE!!!

Mathematics

2 answers:

Marina CMI [18]2 years ago

8 0

Answer:

The maximum profit of $ 10277.32____ can be made when the selling price of the dog food is set to $ _34___ per bag.

Step-by-step explanation:

Profit = Revenue - Cost

P(x) = R(x) -C(x)

= -31.72x^2 + 2,030x -( -126.96x + 26,391)

Distribute the minus sign

= -31.72x^2 + 2,030x+126.96x - 26,391

Combine like terms

= -31.72 x^2 + 2156.96 x - 26391

This is a parabola. It is facing downwards. The maximum profits is at the vertex ( where the max is)

vertex = h = -b/2a = -(2156.96)/(2*-31.72) = -2156.96/-63.44=34

Evaluate P(x) at x=34 to determine the profit

P(34) = -31.72 (34)^2 + 2156.96 (34) - 26391

-36668.32+73336.64-26391

10277.32

Luba_88 [7]2 years ago

5 0

Answer:

Max profit is: $10,277.32

Selling price: $951.52

Step-by-step explanation:

Profit = Revenue - Cost

Profit = (-31.72x² + 2,030x) - (-126.96x + 26,391)

Profit = -31.72x² + 2156.96x - 26391

= -31.72(x² - 68x + 34² - 34²) - 26391

= -31.72(x² - 2(34)(x) + 34²) + 31.72(34²) - 26391

= -31.72(x - 34)² + 10277.32

Max profit is: $10,277.32

When x = 34

Revenue = -31.72(34²) + 2030(34)

= 32351.68 for 34 bags

Per bag:

32351.68/34

$951.52

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Answer:

We conclude that the equation of the line is:

$y = 5x + 1$

Step-by-step explanation:

The slope-intercept form of the line equation

$y = mx+b$

where

m is the slope

b is the y-intercept

Given the data table

x -3 -5 -7 -9 -11

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From the table taking two points

(-3, -16)

(-5, -26)

Determining the slope between (-3, -16) and (-5, -26)

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(-3,\:-16\right),\:\left(x_2,\:y_2\right)=\left(-5,\:-26\right)$

$m=\frac{-26-\left(-16\right)}{-5-\left(-3\right)}$

$m=5$

Thus, the slope of the line is: m = 5

substituting m = 5 and (-3, -16) in the slope-intercept form of the line equation to determine the y-intercept b

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b = -16+15

b = 1

Thus, the y-intercept b = 1

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2 years ago

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natima [27]

Answer:

a) 0.05

b) 0.9826

c) 0.000039308

Step-by-step explanation:

a) $P_X(0) = \frac{e^{-3}3^0}{0!} = e^{-3} = 0.05$

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$P_X(0) = e^{-6}$

$P_X(1) = \frac{e^{-6}*6^1}{1!} = 6*e^{-6}$

Hence,

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$P_X(0) = e^{-15}$

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$P_X(2) = \frac{e^{-15}*15^2}{2!} = 112.5*e^{-15}$

As a result,

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