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faust18 [17]
4 years ago
9

If the volume of the pyramid shown is 108 inches cubed, what is the area of its base?

Mathematics
2 answers:
julia-pushkina [17]4 years ago
8 0

The question is missing the figure. So, the figure is attached below.

Answer:

The area of the base of the pyramid is <u>36 square inches.</u>

Step-by-step explanation:

Given:

Volume of the pyramid is, V=108\ in^3

The height of the pyramid is, h=9\ in

Let the area of the base be 'A'.

So, the volume of the pyramid is given as:

V=\frac{1}{3}Ah

Rewrite the given formula in terms of 'A'. This gives,

A=\frac{3V}{h}

Now, plug in 108 for 'V', 9 for 'h' and solve for 'A'. This gives,

A=\frac{3\times 108}{9}\\\\A=3\times 12\\\\A=36\ in^2

Therefore, the area of the base of the pyramid is 36 square inches.

guajiro [1.7K]4 years ago
5 0

Answer:

D - 36

Step-by-step explanation:

just took the test on edg 2020

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4 years ago
In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, u
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Answer:

      \large\boxed{\large\boxed{0.119}}

Explanation:

You need to find the probability that exactly three of the first 11 inspected packages are damaged and the fourth is damaged too.

<u>1. Start with the first 11 inspected packages:</u>

a) The number of combinations in which 11 packages can be taken from the 20 available packages is given by the combinatory formula:

     C(m,n)=\dfrac{m!}{m!(m-n)!}

      C(20,11)=\dfrac{20!}{11!\cdot(20-11)!}

b) The number of combinations in which 3 damaged packages can be chossen from 7 damaged packages is:

      C(7,3)=\dfrac{7!}{3!\cdot(7-3)!}

c) The number of cominations in which 8 good packages can be choosen from 13 good pacakes is:

      C(13,8)=\dfrac{13!}{8!\cdot(13-8)!}

d) The number of cominations in which 3 damaged packages and 8 good packages are chosen in the first 11 selections is:

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e) The probability is the number of favorable outcomes divided by the number of possible outcomes, then that is:

        \dfrac{C(7,3)\times C(13,8)}{C(20,11)}

Subsituting:

             \dfrac{\dfrac{7!}{3!\cdot(7-3)!}\times \dfrac{13!}{8!\cdot(13-8)!}}{\dfrac{20!}{11!\cdot(20-11)!}}

             =\dfrac{\dfrac{7!}{3!\cdot 4!}\times \dfrac{13!}{8!\cdot 5!}}{\dfrac{20!}{11!\cdot 9!}}=0.26818885

<u>2. The 12th package</u>

The probability 12th package is damaged too is 7 - 3 = 4, out of 20 - 11 = 9:

<u>3. Finally</u>

The probability that exactly 12 packages are inspected to find exactly 4 damaged packages is the product of the two calculated probabilities:

         0.26818885\times 4/9=0.119

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