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Marina CMI [18]
3 years ago
15

Need Assistance With This Problem​

Mathematics
2 answers:
AnnZ [28]3 years ago
8 0

Answer:

not sure how to really answer this question.

NNADVOKAT [17]3 years ago
6 0

Answer:

4.56,  4.65, 5.46, 5.64, 6.45, 6.54

Step-by-step explanation:

First we have to compare the first digits in each number as less is this digit as less is the number. So the least off all are

4.56 and 4.65

which of these two numbers is least ? Now we have to look to the 2-nd digits of these numbers:

they are 5 and 6 . 5<6 so 4.56<4.65

Lets select next numbers whicj first digit is 5. They are:

5.46 and 5.64. However the second digit of the number 5.64 -6 is bigger than the second digit of number 5.46 -4. That is why 5.46<5.64

Similarly 6.45< 6.54

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sergij07 [2.7K]

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Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.
oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

H_0: \mu=2.4\\\\H_a:\mu> 2.4

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58

The degrees of freedom for this sample size are:

df=n-1=45-1=44

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.5801)=0.0066

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0
3 years ago
:Phil’s age decreased by 9 years is 14 years ?
Crank

Answer:

Phil is 23

Step-by-step explanation:

Add 14 to 9 and you get 23. So then you subtract 9 from 23

8 0
2 years ago
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