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11111nata11111 [884]
2 years ago
9

Can someone help me out with this problem How tall is a stack of cube shaped blocks whose volumes are 375 cube inches, 648 cube

inches, and 1029 cube inches?
Mathematics
1 answer:
S_A_V [24]2 years ago
7 0

Answer:

<em>The stack of blocks is 25.96 inches tall</em>

Step-by-step explanation:

<u>The Volume of a Cube</u>

Given a cube-shaped figure of side length L, the volume is calculated by:

V=L^3

If the volume is known, then the side length can be found by solving for L:

L=\sqrt[3]{L}

We are given three cube-shaped blocks with known volumes. We'll find the individual side lengths and add them up to find the height that results when they stack up vertically.

Block 1, volume 375 cube inches:

L_1=\sqrt[3]{375}=7.21\ inch

Block 2, volume 648 cube inches:

L_2=\sqrt[3]{648}=8.65\ inch

Block 3, volume 1029 cube inches:

L_3=\sqrt[3]{1029}=10.10\ inch

Total height=7.21 inch+8.65 inch+10.10 inch=25.96 inch

The stack of blocks is 25.96 inches tall

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The answer is around 27.63 seconds.

Step-by-step explanation:

1 km equals 1000 meters so we have to multiply 10,500 and 1,000 which would equal 10,500,000 km. 10,500,000 divided by 380,000 which is around 27.63 seconds.

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Step-by-step explanation:

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Answer:The countries with lower GDP are the same as those with the lowest HDI.

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Step-by-step explanation:

Data Given are as follows.

Truck arrival rate is given by,   α  = 36 / day

Truck operation departure rate is given,   β= 48 / day

A constructed queuing model is such that so that queue lengths and waiting time can be predicted.

In queuing theory, we have to achieve economic balance between number of customers arriving into system and that of leaving the system whether referring to people or things, in correlating such variables as how customers arrive, how service meets their requirements, average service time and extent of variations, and idle time.

This problem is solved by using concept of Single Channel Arrival with exponential service infinite populate model.

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        where w_s is waiting time in system

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w_{s} = \frac{1}{\alpha - \beta  }

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w_{s} = \frac{1}{12 } day

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Therefore, time required for waiting in system is 2 hours.

           

                   

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