What is the median of the ages 17,16,13,12,10

State the converse<br><br> Please helps asap

Paladinen [302]

Answer: True

Step-by-step explanation:

If x2 = 49, then x = 7. True; x = 7 if and only if x^2 = 49.

I hope this solves your problem! It was a little tricky but I tried.

please mark me Brainliest if your able

have a blessed day

7 0

3 years ago

What is the factors of 2x + 4

ra1l [238]

Answer:

2(x+2)

Step-by-step explanation:

1) 2 : the greatest common factor

2) 2x+4

2x+4=2 ⋅ 2 : Separate out 2

Hope that helps

3 0

3 years ago

Kate has 787 pennies in her piggy bank. She earns 292 pennies by

ivolga24 [154]

the total number of pennies Kate's have can be known by adding all the pennies.

=787 + 292

=1079

the nearest round off of the pennies is 1000

3 0

3 years ago

Read 2 more answers

The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However,

strojnjashka [21]

Answer:

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

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".

$\bar X=400$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=80$ represent the population standard deviation

n=15 represent the sample size

2) Part a

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

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

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: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

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

$400-2.58\frac{80}{\sqrt{15}}=346.708$

$400+2.58\frac{80}{\sqrt{15}}=453.292$

So on this case the 99% confidence interval would be given by (346.708;453.292)

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

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

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

We need to find the degrees of freedom first df=n-1=15-1=14

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(0.005,14)".And we see that $t_{\alpha/2}=2.98$

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

$400-2.98\frac{80}{\sqrt{15}}=338.445$

$400+2.98\frac{80}{\sqrt{15}}=461.555$

So on this case the 99% confidence interval would be given by (338.445;461.555)

4 0

3 years ago

Find the interquartile range of the data.

likoan [24]

The answer is 14. Here is how I work it out.

First we are going have to identify quartile 1 and quartile 3.

So after putting the numbers in order from least to greatest mark the number with a half way point.

This is optional but it will help us spot the sections better.

31,33,35,41,43,|46,48,49,49,50

Next put parentheses around the remaining groups of numbers.

(31,33,35,41,43,)|(46,48,49,49,50)

For the next step we have to find the median of each group.

(31,33,35,41,43,)|(46,48,49,49,50)

The median of each group, are called the quartiles, the median of the lower half is quartile 1, and the median of the upper half is quartile 3.

Now subtract quartile 1 from quartile 3.

49 – 35 = 14

So the interquartile range of this data set is 14.

4 0

3 years ago