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2 years ago

If you have a population standard deviation of 9 and a sample size of 144, what is your standard error of the mean?

Mathematics

2 answers:

andrezito [222]2 years ago

6 0

So your answer would be 1296

Anettt [7]2 years ago

4 0

Answer: 0.75

Step-by-step explanation:

Standard error is the quotient of the standard deviation and the square root of the sample size. From the question, Standard error= 9/√144= 3/4 or 0.75

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Mrs. Martini is 6 years more than 3 times as old as Tony. Eight years ago, she

zalisa [80]

Mrs Martini age is 42 years and Tony age is 12 years

Step-by-step explanation:

The given is:

Mrs. Martini is 6 years more than 3 times as old as Tony
Eight years ago, she was 2 years less than 9 times as old as he was then

Assume that Tony is x years old now

∵ Mrs. Martini is 6 years more than 3 times as old as Tony

∵ Tony is x years old

- Multiply the age of Tony by 3 and add 6 to the product

∴ The age of Mrs Martini = 3 x + 6

8 years ago

- Subtract 8 from the age of each one

Tony age = (x - 8)

Mrs Martini age = (3 x + 6) - 8 = 3 x + 6 - 8 = 3 x - 2

∵ Eight years ago, Mrs Martini was 2 years less than 9 times as old

as Tony was then

- Multiply Tony age by 9 and subtract 2 from the product

∵ Tony age from 8 years ago = x - 8

∵ Mrs Martini age from 8 years ago = 3 x - 2

∴ 9(x - 8) - 2 = 3 x - 2

- Simplify the left hand side

∴ 9 x - 72 - 2 = 3 x - 2

- Add like terms

∴ 9 x - 74 = 3 x - 2

- Subtract 3 x from both sides

∴ 6 x - 74 = -2

- Add 74 to both sides

∴ 6 x = 72

- Divide both sides by 6

∴ x = 12

∵ x represents the age of Tony

∴ Tony is 12 years old

∴ 3 x + 6 is the age of Mrs Martini

- Substitute x by 12

∵ 3(12) + 6 = 36 + 6 = 42

∴ Mrs Martini is 42 years old

Mrs Martini age is 42 years and Tony age is 12 years

Learn more:

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Which of the following options is equivalent to 140%? Select all that apply.

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The total arm and blade of a windshield wiper is 12 in. long and rotates back and forth through an angle of 85°. The shaded regi

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The amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation. Ric

Papessa [141]

Let $X$ denote the number of minutes spent brushing his teeth, and let $\mu$ be the mean and $\sigma$ the standard deviation for this distribution.

$\mathbb P(X$

The z-score corresponding to this probability is approximately $z=-0.2533$ , which means

$\dfrac{1-\mu}\sigma=-0.2533\iff\mu-0.2533\sigma=1$

Next, (note the sign change)

$\mathbb P(X>2)=0.02\implies\mathbb P(X\le2)=\mathbb P\left(\dfrac{X-\mu}\sigma\le\dfrac{2-\mu}\sigma\right)=0.98$

The corresponding z-score is approximately $z=2.0538$ , so you have

$\dfrac{2-\mu}\sigma=2.0538\iff\mu+2.0538\sigma=2$

Solving the two equations for $\mu$ and $\sigma$ , you'll find that the mean is approximately $\mu=1.1098$ and the standard deviation is approximately $\sigma=0.4334$ .

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2 years ago