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kap26 [50]
3 years ago
7

Please Help!!! I have 5 minutes left!!!!! The volume of a​ cone-shaped hole is 16 pi ft cubed . If the hole is 3 ft​ deep, what

is the radius of the​ hole?
Mathematics
2 answers:
tangare [24]3 years ago
8 0

Answer:

3.0699

Step-by-step explanation:

V = pi*r^2*h

you have v and h and pi so you can get r

postnew [5]3 years ago
5 0

Answer:

V = 1/3 • π • radius cubed • height

If the volume is is 16π and the height is 3, then 16π = 1/3 • π • r cubed • 3

16π ÷ (1/3 • π • 3) = 16

radius squared = 16        √16 = 4, so radius = 4

You can also check: 1/3 • r cubed • 3 = 16    The only value left to multiply is π

Step-by-step explanation:

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IRISSAK [1]

Answer:

a) 0.9

b) Mean = 1.58

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Step-by-step explanation:

We are given the following in the question:

A marketing firm is considering making up to three new hires.

Let X be the variable describing the number of hiring in the company.

Thus, x can take values 0,1 ,2 and 3.

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P(x\geq 2) = P(x=2) + P(x=3)\\0.5 = P(x=2) + 0.18\\ P(x=2) = 0.32

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P(x= 0) +P(x= 1) + P(x= 2) + P(x= 3) = 1\\ 0.1 + P(x= 1) + 0.32 + 0.18 = 1\\ P(x= 1) = 1- (0.1+0.32+0.18) = 0.4

\text{P(firm will make at least one hire)}\\= P(x\geq 1)\\=P(x=1) + P(x=2) + P(x=3)\\ = 0.4 + 0.32 + 0.18 = 0.9

b) expected value and the standard deviation of the number of hires.

E(X) = \displaystyle\sum x_iP(x_i)\\=0(0.1) + 1(0.4) + 2(0.32)+3(0.18) = 1.58

E(x^2) = \displaystyle\sum x_i^2P(x_i)\\=0(0.1) + 1(0.4) + 4(0.32) +9(0.18) = 3.3\\V(x) = E(x^2)-[E(x)]^2 = 3.3-(1.58)^2 = 0.80\\\text{Standard Deviation} = \sqrt{V(x)} = \sqrt{0.8036} = 0.89

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Step-by-step explanation:

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