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musickatia [10]
3 years ago
14

Number 17 (the circled one)

Mathematics
1 answer:
olga nikolaevna [1]3 years ago
3 0
It would take 15 hours.
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Which number is a perfect cube? 5, 100, 125, 150​
KATRIN_1 [288]

Answer:

125

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation
fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 197.5, \sigma = 8.3

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

Z = \frac{210.9 - 197.5}{8.3}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

Z = \frac{196 - 197.5}{2.14}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0
3 years ago
An appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W. From pre
melamori03 [73]

Answer:

We conclude that a compact microwave oven consumes a mean of more than 250 W.

Step-by-step explanation:

We are given that an appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W with a population standard deviation of 15 W.

They take a sample of 20 microwave ovens and find that they consume a mean of 257.3 W.

Let \mu = <u><em>mean power consumption for microwave ovens.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 250 W     {means that a compact microwave oven consumes a mean of no more than 250 W}

Alternate Hypothesis, H_A : \mu > 250 W     {means that a compact microwave oven consumes a mean of more than 250 W}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                                T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean power consumption for ovens = 257.3 W

            σ = population standard deviation = 15 W

            n = sample of microwave ovens = 20

So, <em><u>the test statistics</u></em>  =  \frac{257.3-250}{\frac{15}{\sqrt{20} } }

                                      =  2.176

The value of z test statistics is 2.176.

<u>Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.</u>

Since our test statistic is more than the critical value of t as 2.176 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that a compact microwave oven consumes a mean of more than 250 W.

7 0
3 years ago
A bakery uses 25 cups of flour to make 8 loaves of bread. What is true about the number of cups of flour in each loaf?
Levart [38]
So basically we need to divide

25/8=3.125

they need 3.125 cups of flour in each loaf

I hope I've helped!
6 0
3 years ago
Help me out please..........
evablogger [386]
I’m not sure for the other ones but D is def. likely
7 0
3 years ago
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