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MissTica
3 years ago
14

Check my steps and for part b, the professor wants us to show are work and I don't know how to solve it by hand.

Mathematics
1 answer:
Ugo [173]3 years ago
4 0
Part a is correct.

Part b can be done a couple ways.
1) Use binomial distribution. p = 0.2, n = 144
P(x>3) = 1 - P(x=0) -p(x=1) -p(x=2) -p(x=3)
where
P(x=k) = (nCk)p^k (1-p)^{n-k}

2) Assume a normal distribution with mean = 28.2, stdev = 4.8
z = \frac{4 - 28.2}{4.8} = -5.04
Look up z-value in normal table to find probability.

Hope that helps.
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A rectangular prism has a base with a length of 25, a width of 9, and a height of 12. A second prism has a square base with a si
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Answer:

The height of the second prism is 12 units.

Step-by-step explanation:

First, we must know that the volume of a rectangular prism is equal to length x width x height. With this information, we can multiply the given dimensions of the first rectangular prism to get a volume of 2700 cubic units. From here, we must understand wha we need from the second rectangular prism. Because it has a square base of 15, that accounts for both its width and length. Therefore, we have that

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2 years ago
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3 years ago
A roll of steel is manufactured on a processing line. The anticipated number of defects in a 10-foot segment of this roll is two
Vikentia [17]

Answer:

the probability of no defects in 10 feet of steel = 0.1353

Step-by-step explanation:

GIven that:

A roll of steel is manufactured on a processing line. The anticipated number of defects in a 10-foot segment of this roll is two.

Let consider β to be the average value for defecting

So;

β = 2

Assuming Y to be the random variable which signifies the anticipated number of defects in a 10-foot segment of this roll.

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Y \sim P( \beta = 2)

the probability mass function can be represented as follows:

\mathtt{P(y) = \dfrac{e^{- \beta} \ \beta^ \ y}{y!}}

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y =  0,1,2,3 ...

Hence,  the probability of no defects in 10 feet of steel

y = 0

\mathtt{P(y =0) = \dfrac{e^{- 2} \ 2^ \ 0}{0!}}

\mathtt{P(y =0) = \dfrac{0.1353  \times 1}{1}}

P(y =0) = 0.1353

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3 years ago
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