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frozen [14]
3 years ago
15

A production manager randomly sampled production lines at a factory that produces automobiles. She wanted to find out how many p

roduction lines caused defects in newly produced automobiles. The proportion of production lines that caused defects was 0.08, with a margin of error of 0.01. Construct a confidence interval for the proportion of production lines that caused defects
Mathematics
1 answer:
Elena-2011 [213]3 years ago
5 0

Answer:

The confidence interval for the proportion of production lines that caused defects is (0.07, 0.09).

Step-by-step explanation:

A confidence interval for a population proportion is a function of the sample proportion and the margin of error.

The interval has two bounds, a lower bound and an upper bound.

The lower bound is the sample proportion subtracted by the margin of error.

The upper bound is the margin of error added to the sample proportion.

In this problem, we have that:

Sample proportion 0.08

Margin of error 0.01

0.08 - 0.01 = 0.07

0.08 + 0.01 = 0.09

The confidence interval for the proportion of production lines that caused defects is (0.07, 0.09).

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Answer:

There are 15 combinations.

Step-by-step explanation:

A restaurant is offering a dinner special that includes one starter and one entree.

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So, we have 3 starters and 5 entrees.

To know the possible dinner special combinations we will simply multiply the two.

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Answer:

58.9% produced produced peppers weighing between 13 and 16 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 15

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Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

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P(13 \leq x \leq 16) = P(\displaystyle\frac{13 - 15}{1.75} \leq z \leq \displaystyle\frac{16-15}{1.75}) = P(-1.142\leq z \leq 0.571)\\\\= P(z \leq 0.571) - P(z < -1.142)\\= 0.716 - 0.127 = 0.589 = 58.9\%

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Answer:

c. m∠2 = m∠5

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