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mezya [45]
2 years ago
6

Product A requires 5 machine hours per unit to be produced, Product B requires only 3 machine hours per unit, and the company's

productive capacity is limited to 240,000 machine hours. Product A sells for $16 per unit and has variable costs of $6 per unit. Product B sells for $12 per unit and has variable costs of $5 per unit. Assuming the company can sell as many units of either product as it produces, the company should:
Produce only Product A.
Produce only Product B.
Produce equal amounts of A and B.
Produce A and B in the ratio of 62.5% A to 37.5% B.
Produce A and B in the ratio of 40% A and 60% B.
Mathematics
1 answer:
Lesechka [4]2 years ago
5 0

Answer:

  Produce only Product B.

Step-by-step explanation:

The contribution margin per machine hour for product A is ...

  ($16 -$6)/(5 hour) = $2 per hour

The contribution margin per machine hour for product B is ...

  ($12 -$5)/(3 hour) ≈ $2.33 per hour

The company should produce the maximum possible number of the product that contributes the most per machine hour: Product B.

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If four is half of five and two thirds of six is nine. what is thirty-two?
valkas [14]
Four is IV which is half of five. IX is nine which is 2 thirds of 6 so... thirty-two is 1 5/6
5 0
2 years ago
Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to
RUDIKE [14]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

5 0
2 years ago
Gianna is going to throw a ball from the top floor of her middle school. When she throws the hall from 48 feet above the ground,
vazorg [7]

Answer:

So, the times the ball will be 48 feet above the ground are t = 0 and t = 2.

Step-by-step explanation:

The height h of the ball is modeled by the following equation

h(t)=-16t^2+32t+48

The problem want you to find the times the ball will be 48 feet above the ground.

It is going to be when:

h(t) = 48

h(t)=-16t^{2}+32t+48

48=-16t^{2}+32t+48

0=-16t^{2}+32t+48 - 48

16t^{2} - 32t = 0

We can simplify by 16t. So

16t(t-2)= 0

It means that

16t = 0

t = 0

or

t - 2 = 0

t = 2

So, the times the ball will be 48 feet above the ground are t = 0 and t = 2.

6 0
2 years ago
Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

b)the length of the wire for the square = (\frac{240}{\pi+4}) in

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=\frac{y}{4}

Area of the square which is L² can now be said to be;

A_S=(\frac{y}{4})^2 = \frac{y^2}{16}

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

A_C= \pi (\frac{60-y}{2\pi } )^2

     =( \frac{60-y}{4\pi } )^2

Total Area (A);

A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0
2 years ago
In 10 minutes, Max ran 1.45 km while Avery ran 1.61 km. How much more did<br> Avery run?
SOVA2 [1]

Answer:

0.16km

Step-by-step explanation:

Unless I am missing something it is a simple subtraction problem and the time is not needed. It is simply asking how much farther Avery ran than Max, which is 1.61km - 1.45km which is 0.16km

6 0
1 year ago
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