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shepuryov [24]
3 years ago
10

What's 4.88 rounded to the nearest tenth......

Mathematics
2 answers:
maria [59]3 years ago
7 0
4.90

The 4 in 4.88 is in the ones place. Then the first 8 is in the tenths place and the second 8 is in the hundreths place. Now forget the 4 for now and just treat the .88 like 88. Round it up to the nearest ten place wich would be 90. Now put back the for and turn the 90 into a decimal. Then you will have your awnser.
Ugo [173]3 years ago
5 0
The first decimal point is tenth. 4.88
when rounding, we just look at the number right after the number you want to round to. it's 8 right? since 8 is closer to 10 than 0,
we round it to 4.9
                  (4.8 + 0.10)
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Answer:

When sample size increases from n = 23 to n = 35, the mean of the distribution of sample means remains same but the standard deviation of the distribution of sample means decreases from 0.073 to 0.059.

Step-by-step explanation:

We are given the amounts of time employees at a large corporation work each day are normally distributed, with a mean of 7.5 hours and a standard deviation of 0.35 hour.

Random samples of size 23 and 35 are drawn from the population and the mean of each sample is determined.

Here, \mu = population mean = 7.5 hours

          \sigma = population standard deviation = 0.35 hour

<u><em /></u>

<u><em>Let </em></u>\bar X<u><em> = sample mean</em></u>

(a) The random samples of size of n = 23 is drawn from the population;

So, mean of the distribution of sample means = \mu = 7.5 hours

Standard deviation for the distribution of sample means is given by;

               s  =  \frac{\sigma}{\sqrt{n} } = \frac{0.35}{23} = 0.073

(b) The random samples of size of n = 35 is drawn from the population;

So, mean of the distribution of sample means = \mu = 7.5 hours

Standard deviation for the distribution of sample means is given by;

               s  =  \frac{\sigma}{\sqrt{n} } = } \frac{0.35}{\sqrt{35} } = 0.059

(c) So, as we can see that when sample size increases from n = 23 to n = 35, the mean of the distribution of sample means remains same but the standard deviation of the distribution of sample means decreases from 0.073 to 0.059.

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3 years ago
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Claim: The hypertension medicine lowered blood pressure.

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Sample size = n = 13

Sample mean of difference = \bar{d} = 10.1

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Test statistic is

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Degrees of freedom = n - 1 = 13 - 1 = 12

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Test statistic | t | > critical value we reject null hypothesis.

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