13,501 rounded to the nearest 10,000

2 answers:

7 0

You look at the number just to the right of the 10000s place, in this case 3, if it is less than five you round everything after the 10000s place down to zero, otherwise you round up the 10000s place and everything else to zero...

In this case 3 follows the 10000s place it becomes:

10000

galben [10]3 years ago

6 0

It would be 10,000 because if the number behind it is 4 or under it goes down if its 5 and up it goes up. even if you rounded 3501 to 4000 you would stilll get 10,000

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Suppose that the sample standard deviation was s = 5.1. Compute a 98% confidence interval for μ, the mean time spent volunteerin

NISA [10]

Answer:

The 95% confidence interval would be given by (5.139;5.861)

Step-by-step explanation:

1) 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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

2) Confidence interval

Assuming that $\bar X =5.5$ and the ranfom sample n=1086.

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:

$df=n-1=1086-1=1085$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,1085)".And we see that $t_{\alpha/2}=2.33$

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

$5.5-2.33\frac{5.1}{\sqrt{1086}}=5.139$

$5.5+2.333\frac{5.1}{\sqrt{1086}}=5.861$

So on this case the 95% confidence interval would be given by (5.139;5.861) We are 98% confident that the mean time spent volunteering for the population of parents of school-aged children is between these two values.

5 0

3 years ago

I need help quick plz

Bess [88]

Hello, the answer is C

7 0

4 years ago

Read 2 more answers

What is the answer of 10 to the power of -4<br><br> [tex] 10^-4

malfutka [58]