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2 years ago

11

Which of these is a statistical question? I accidentally clicked an clicked an answer, but I want to know which one is the right

one.

1 answer:

sasho [114]2 years ago

7 0

Step-by-step explanation:

I think A is the answer to your question

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The gcf of two numbers less than or equal to 12 is 2. Their LCM is 20.What are the numbers?

bixtya [17]

We know:

$LCM(a,\ b)=\dfrac{ab}{GCF(a,\ b)}$

We have

$GCF(a,\ b)=2,\ LCM(a,\ b)=20,\ a\leq12,\ b\leq12$

Substitute:

$20=\dfrac{ab}{2}\qquad|\cdot2\\\\ab=40$

$40=1\cdot40=2\cdot20=4\cdot10=5\cdot8$

Only 4 and 10 meet the requirements of the question.

--------------------------------------------------------------------

40 and 1 NOT, because 40 > 12

20 and 2 NOT, because 20 > 12

5 and 8 NOT because GCF(5, 8) = 1

5 0

2 years ago

The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

Alexandra [31]

Answer:

$\bar x = 260.1615$

$\sigma = 70.69$

The confidence interval of standard deviation is: $53.76$ to $103.25$

Step-by-step explanation:

Given

$n =20$

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

$\bar x = 260.1615$

$\bar x = 260.16$

Solving (b): The standard deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

So, the confidence interval of the standard deviation is:

$\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} }$ to $\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }$

$70.69 * \sqrt{\frac{20 - 1}{32.852}$ to $70.69 * \sqrt{\frac{20 - 1}{8.907}$

$70.69 * \sqrt{\frac{19}{32.852}$ to $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

8 0

2 years ago

A bag contains three red marbles and two blue marbles. Marcus reaches in and pulls out a marble, replaces it, but then reaches i

Svetradugi [14.3K]

Answer:

your answer would be $\frac{1}{5}$

Hope this helps! (:

5 0

3 years ago

Wanna do meet do this code and click ask to join <br> afg-wrkj-uud

Zinaida [17]

Answer:

you talking to specific people or can anyone join

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

WILL MARK THE CORRECT ANSWER BRAINLIEST!<br><br> Select all the correct answers:

stepladder [879]

Answer:

B, D

Step-by-step explanation:

Solve all equations:

A.

$x-3x=-2x+4\\ \\-2x=-2x+4\\ \\0=4$

This is false equality, so this equation has no solutions.

B.

$x-\dfrac{1}{3}(3x)=0\\ \\x-x=0\\ \\0=0$

This equality is true for all values of x, so this equation has infinitely many solutions.

C.

$\dfrac{1}{3}x-3=4+2x\\ \\x-9=12+6x\\ \\x-6x=12+9\\ \\-5x=21\\ \\x=-\dfrac{21}{5}$

This equation has the unique solution.

D.

$3x+2-\dfrac{1}{3}(3x)=2+2x\\ \\9x+6-3x=6+6x\\ \\6x+6=6+6x\\ \\0=0$

This equality is true for all values of x, so this equation has infinitely many solutions.

E.

$x-\dfrac{1}{3}x+2=0\\ \\3x-x+6=0\\ \\2x=-6\\ \\x=-3$

This equation has the unique solution.

8 0

2 years ago