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3 years ago

(9.2×10^5)(4×10^-3)

Mathematics

1 answer:

Len [333]3 years ago

5 0

first change the 9.2 to standard form. which is

$92 \times {10}^{ - 1}$

now the whole equation becomes

$(92 \times {10}^{ - 1} \times {10}^{ - 5} ) \times (4 \times {10}^{ - 3} )$

$= 92 \times 4 \times {10}^{ - 1 + 5 - 3}$

$= 368 \times {10}^{1}$

$= 3.68 \times {10}^{2 + 1}$

$= 3.68 \times {10}^{3}$

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3 years ago

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3 years ago

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3 years ago

If Upper X overbar equals 62, Upper S equals 8, and n equals 36, and assuming that the population is normally distributed, c

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Answer:

The 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=62$ represent the sample mean

$\mu$ population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size

Part a: Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=36-1=35$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,35)".And we see that $t_{\alpha/2}=2.72$

Now we have everything in order to replace into formula (1):

$62-2.72\frac{8}{\sqrt{36}}=58.373$

$62+2.72\frac{8}{\sqrt{36}}=65.627$

So on this case the 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

3 0

3 years ago

(9.2×10^5)(4×10^-3)​

(9.2×10^5)(4×10^-3)