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Ludmilka [50]
2 years ago
6

A delivery truck company just bought a new delivery truck and they need to know the maximum volume it can carry. In the front of

the truck, there is an extra ledge that sticks out over the driver's cab for extra storage space. What is the maximum amount of cargo that can fit into the new truck?

Mathematics
1 answer:
sergejj [24]2 years ago
4 0

Answer:

The answer is below

Step-by-step explanation:

To find the maximum amount of cargo the truck can carry, we need to find the volume of the truck.

Volume = length × width × height.

Firstly 1 feet (1') = 12 inches (12"),

For the extra ledge that sticks out, the height = 7'8" = 7.667 feet, the width = 16'9" - 14'3" = 16.75 - 14.25 = 2.5 feet, the length = 2'7" = 2.583 feet

Volume of extra ledge = length × width × height = 2.583 × 2.5 × 7.667 = 49.5 feet³

For the truck, the height = 7'8" = 7.667 feet, the length = 14'3" = 14.25 feet, the width = 6'6" = 6.5 feet

Volume of truck = length × width × height = 14.25 × 6.5 × 7.667 = 710.16 feet³

The maximum volume = volume of extra ledge + volume of truck = 49.5 + 710.16 = 759.66 feet³

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Ava wants to figure out the average speed she is driving. She starts checking her car’s clock at mile marker 0. It takes her 4 m
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<h2>Answer:</h2>

The average speed of car is:

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<h2>Step-by-step solution:</h2>

The table that describes the time and number of  miles marked is given by:

Time       Miles

  0              0

  4               3

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Answer:

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Step-by-step explanation:

Jar A contains 4*0.45 L acid, and 4 L of a solution  of acid.

Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.

Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.

50% = 0.5

For jar A.

(2/3)*k/100 L acid  is added to jar A.

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We also can find k using jar B.

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Check.

We can check at least for jar A.

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2/3 L of the solution from jar C was added, and now we have 4 2/3 L of solution.

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L solute/L solution =  [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50%  as it is given that jar A has 50% at the end.

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Since population standard deviation is unknown, so we use t-test.

Critical value for  95 percent confidence interval  :

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Confidence interval : \overline{x}\pm t_{n-1, \alpha/2}\dfrac{s}{\sqrt{n}}

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Required 95% confidence interval :  (15.263,\ 17.537)

8 0
2 years ago
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