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kati45 [8]
2 years ago
8

Round 13,046 to the nearest thousand.

Mathematics
2 answers:
Margaret [11]2 years ago
8 0

Answer:

13,000

Step-by-step explanation:

When rounding, first:

Find the place value that you are looking for, in this case, being the thousands place value. It is a 3:

1<u>3</u>,046

When you round, you will look at the place value directly next to the one you are trying to find (which is the thousands place value). In this case, it will be the hundreds place value, which has the value of 0.

When the value of the number is 5 or greater, you round up.

When the value of the number is 4 or less, you round down.

In this case, it is 0, which is less than 4, so you round down.

13,046 to the nearest thousand place value is 13,000.

~

klemol [59]2 years ago
7 0

Answer:

13,000

Step-by-step explanation:

13,046 ,1=ten thousand,3=thousand,0=hundred,4=tens,6=units

the number after the 3 is 0 which is less then five and can't be rounded up to one

therefore the answer is 13,000

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(a) Null Hypothesis, H_0 : \mu \leq 41  

    Alternate Hypothesis, H_A : \mu > 41

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Step-by-step explanation:

We are given that in a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35.

It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41.

Let \mu = <u><em>mean sunspot activity during the Spanish colonial period.</em></u>

(a) Null Hypothesis, H_0 : \mu \leq 41     {means that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41}

Alternate Hypothesis, H_A : \mu > 41     {means that the mean sunspot activity during the Spanish colonial period was higher than 41}

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 = 47

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            n = sample of periods from Spanish colonial times = 40

So, <em><u>the test statistics</u></em>  =  \frac{47-41}{\frac{35}{\sqrt{40} } }

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(b) The value of z test statistics is 1.08.

(c) <u>Now, the P-value of the test statistics is given by;</u>

                P-value = P(Z > 1.08) = 1 - P(Z < 1.08)

                              = 1 - 0.8599 = <u>0.1401</u>

Since, the P-value of the test statistics is higher than the level of significance as 0.1401 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

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Step-by-step explanation:

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