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4vir4ik [10]
2 years ago
11

A paper company needs to ship paper to a large printing business. The paper will be shipped in small boxes and large boxes. The

volume of each small box is 7 cubic feet and the volume of each large box is 13 cubic feet. There were twice as many large boxes shipped as small boxes shipped and the total volume of all boxes was 165 cubic feet. Determine the number of small boxes shipped and the number of large boxes shipped.
Mathematics
1 answer:
Molodets [167]2 years ago
6 0

Answer:

small boxes = 5 and large boxes = 10

Step-by-step explanation:

Given:

The volume of each small box is 7 cubic feet.

volume of each large box is 13 cubic fee.

There were twice as many large boxes shipped as small boxes shipped.

Total volume of all boxes was 165 cubic feet.

Question asked:

The number of small boxes shipped and the number of large boxes shipped?

Solution:

Let  number of small boxes shipped = x

Then, the number of large boxes shipped = 2x (given)

Total volume of all boxes = 165 cubic feet.

no. of small boxes \times vol. of each small box +  no. of large boxes \times vol. of each large box =  165 cubic feet. (Total volume of all boxes)

x \times7 + 2x \times 13 = 165\\7x + 26x = 165\\33x = 165

Dividing both side by 33

x = \frac{165}{33}\\ x = 5

The number of small boxes shipped = x = 5

The number of large boxes shipped = 2 x = 2\times 5 = 10

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mariarad [96]

Answer:

a. the probability that any one of the computers will require repair on a given day is constant

Step-by-step explanation:

The following properties must be true in order for a distribution to be binomial:

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There are only two outcomes (requires repair or do not require repair)

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Xy+(4(20))>x-5y(2+9-7)
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I'll solve for y xy+(4(20))>x-5y(2+9-7) 

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3 0
2 years ago
Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u
soldier1979 [14.2K]

x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}

\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

Therefore:

\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]

Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

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The distribution will exhibit symmetry. 
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