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3 years ago

Which of the following are examples of random samples? Select all that apply?

Mathematics

1 answer:

prohojiy [21]3 years ago

3 0

I think all solution apply

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A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not i

LenaWriter [7]

You have a 33.33% chance to get it right each time
multiply 33.33%*33.33%*33.33%= .3333*.3333*.3333
it equals about 4%

6 0

3 years ago

Plz help me. But like for real plz

scoundrel [369]

Answer:

They will pay the same amount!

3 0

3 years ago

Read 2 more answers

12345 what is the answer if you multiply 1x2x3x4x5 and do the same with subtraction, addition and devision, in that order. Then

Alexandra [31]

Well, we just need to perform the operations:

Multiplication: $1\cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120$
Subtraction: $1-2-3-4-5 = -13$
Addition: $1+2+3+4+5 = 15$
Division: $1 \div 2 \div 3 \div 4 \div 5 = \frac{1}{120}$

So, if you add all the numbers together you get

$120-13+15+\dfrac{1}{120} = 122 + \dfrac{1}{120}$

Or, if you prefer,

$\dfrac{1681}{120}$

8 0

3 years ago

Which of the following sets of numbers could represent the three sides of a

Alchen [17]

Answer:

im not sure if this is correct but, i got C

5 0

3 years ago

A marketing researcher wants to find a 96% confidence interval for the mean amount those visitors spend per person per day while

ki77a [65]

Answer:

$n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38$

So then the minimum sample to ensure the condition given is n= 38

Step-by-step explanation:

Notation

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

$\mu$ population mean (variable of interest)

$\sigma=12$ represent the population standard deviation

n represent the sample size

ME = 4 the margin of error desired

Solution to the problem

When we create a confidence interval for the mean the margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 96% of confidence interval now can be founded using the normal distribution. The significance is $\alpha=1-0.96 =0.04$ . And in excel we can use this formula to find it:"=-NORM.INV(0.02;0;1)", and we got $z_{\alpha/2}=2.054$ , replacing into formula (b) we got:

$n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38$

So then the minimum sample to ensure the condition given is n= 38

5 0

4 years ago