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2 years ago

9

Suppose that the probability that an item produced by a certain machine will be defective is 0.1. Find the probability that a sa

mple of 20 items will contain at most 2 defective items.

1 answer:

sesenic [268]2 years ago

4 0

Answer:

no it is really easy question to answered so do it your self

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I need to know how to solve 5(6)+(2)<br> ______<br> 8

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First do 5times 6 equals 30 then add 2 that gives you 32 the next step is to divide by 8 this will give you a final answer of 4

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3 years ago

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Choose exactly two answers that are correct. Which expressions could complete this equation so that it has an infinite number of

Maru [420]

Answers A and <span>D
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</span><span>9x − 3(x+8) [distribute] = 9x − 3x − 24 = 6x − 24 [combine like terms]

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<span>8(x−3)−2x = [distribute] 8x−24−2x = 6x−24 [combine like terms] </span>

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3 years ago

For the data in the table, does y vary directly with x? If it does, write an equation for the direct variation.

QveST [7]

Answer:

d

Step-by-step explanation:

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3 years ago

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The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev

Vladimir [108]

Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 30.05 inches

$\sigma$ = standard deviation = 0.2 inches

n = sample of sheets = 4

So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches) = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } }$ ) = P(Z < 3) = 0.99865

P(X bar <= 30.25) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } }$ ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches) = 0.99865 - 0.97725

= 0.0214

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3 years ago

What is the value of a in this diagram

Delicious77 [7]

What diagram people?

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3 years ago