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3 years ago

Tina babysat eight times. She earned $15, $20, $10, $12, $20, $16, $80, and $18. What is the average amount of money she made?

2 answers:

7 0

To get the average you add up all the numbers then divide by the set of numbers.

The answer I got was: 23.875

round it to what you need.

AleksAgata [21]3 years ago

6 0

The mean is 23.875. That would be roughly $24 each time

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Find the slope of the line between the points (1, 4) and (-1,6)

Oduvanchick [21]

Answer:

slope = - 1

Step-by-step explanation:

Calculate the slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (1, 4) and (x₂, y₂ ) = (- 1, 6)

m = $\frac{6-4}{-1-1}$ = $\frac{2}{-2}$ = - 1

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3 years ago

The "Let's Roll" game uses a number cube with the numbers 2,4,6,8,10, and 12. There are prizes for rolling any number less than

deff fn [24]

There is a 33.3% chance of rolling a number less than 6. It’s 2/6, which is reduced to 1/3. = that to x/100, multiply 1 x 100, and divide that by 3. So it’s 100 divided by 3, and that’s 33.3, which is the percentage.

8 0

3 years ago

Read 2 more answers

Find the area of the figure

GenaCL600 [577]

Find the rectangle area

A = 8*4 = 32 in²

Find the height of trapezoid

8-4 = 4 in

Find the area of trapezoid

(5,3+8)*4/2 = 13,3*4/2 = 53,2/2 = 26,6 in²

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3 years ago

My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the

AVprozaik [17]

Answer:

a) $n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

b) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

Step-by-step explanation:

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by: