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mariarad [96]
3 years ago
9

If John's dog weighs 16.23 lb how much would Jim's dog weigh​

Mathematics
2 answers:
Morgarella [4.7K]3 years ago
6 0

Answer:

It’s 5,41

Step-by-step explanation:

Hunter-Best [27]3 years ago
5 0
Jim's dog would be 5.41lb!!
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Mrs. Harrison borrows $600 to pay for a new phone. The simple interest on the loan is 9% for one year.
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$654

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600*1.09=654

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What is the least common denominator of 4/5 - 3/4
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Step-by-step explanation:

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If C={-10,-9,-8,-7,-6,-5} and D={-6,-5,-4,-3,-3,-1} what is CnD
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3 years ago
Suppose that a random sample of 10 adults has a mean score of 62 on a standardized personality test, with a standard deviation o
aniked [119]

Answer:

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

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The lower end of the interval is the sample mean subtracted by M. So it is 62 - 2.08 = 59.92

The upper end of the interval is the sample mean added to M. So it is 62 + 2.08 = 64.08.

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

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3 years ago
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