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valentina_108 [34]
2 years ago
12

Need help with thisssss

Mathematics
1 answer:
schepotkina [342]2 years ago
8 0

Answer:

4^8

Step-by-step explanation:

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Jakes phone will play music for 12 hours on a full charge . He has been using it for 3 hours . What percent of a full charge is
iogann1982 [59]

Answer:

75%

Step-by-step explanation:

3/12 =1/4

so 1/4 of 100% = 75%

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According to the school survey, 12% of the students at Rockwood Junior High School speak Spanish. There are 36 students at the s
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Mean=np
36=(0.12)n
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300 students were surveyed.
That's the way I answer in my country but I'm not sure about this solution is correct in your school or not.
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A package of fish weighs 2.5 pounds. The price is $3.50 per pound. What is the price of this package?
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Answer:

$8.75

Step-by-step explanation:

If you do

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3 years ago
Exhibit 9-2 The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the cu
Diano4ka-milaya [45]

Answer:

At a .05 level of significance, it can be concluded that the mean of the population is significantly more than 3 minutes.

Step-by-step explanation:

We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

At the null hypothesis, we test if the mean is of at most 3 minutes, that is:

H_0: \mu \leq 3

At the alternative hypothesis, we test if the mean is of more than 3 minutes, that is:

H_1: \mu > 3

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

3 is tested at the null hypothesis:

This means that \mu = 3

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.

This means that n = 100, X = 3.1, \sigma = 0.5

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{3.1 - 3}{\frac{0.5}{\sqrt{100}}}

z = 2

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 3.1, which is 1 subtracted by the p-value of z = 2.

Looking at the z-table, z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228

The p-value of the test is of 0.0228 < 0.05, meaning that the is significant evidence to conclude that the mean of the population is significantly more than 3 minutes.

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3 years ago
A man weekly pay is $x. he spends 1/2 of his pay on food and 1/3 on rent
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