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Luda [366]
3 years ago
8

Please need help asap!!

Mathematics
1 answer:
sergey [27]3 years ago
8 0
Option 2. Is the answer to ur question..!!
Hope this helps you!!

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The mean increases when the outlier is removed
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Is Joan read 74 pages in 4 hours how long will it take her to read 240 pages
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5 0
3 years ago
A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .5
lara [203]

Answer:

7.3% of the bearings produced will not be acceptable

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 0.499, \sigma = 0.002

Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.

So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.

Larger than 0.504

1 subtracted by the pvalue of Z when X = 0.504.

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.504 - 0.494}{0.002}

Z = 2.5

Z = 2.5 has a pvalue of 0.9938

1 - 0.9938= 0.0062

Smaller than 0.496

pvalue of Z when X = -1.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.496 - 0.494}{0.002}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.0668 + 0.0062 = 0.073

7.3% of the bearings produced will not be acceptable

4 0
3 years ago
Find the area of the rectangle ABCD. I need both green and grey box please.
Kitty [74]

Answer:

180

Step-by-step explanation:

12x15=180

5 0
3 years ago
Read 2 more answers
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