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MArishka [77]
3 years ago
11

Last year, Nicholas started babysitting to earn extra spending money. Over the summer, he tracked how much money he earned at ea

ch babysitting jobThis box plot shows the results What percent of the time did Nicholas earn $25 or less?

Mathematics
1 answer:
Ierofanga [76]3 years ago
8 0

Answer:

25%

Step-by-step explanation:

The 5 values of the given box plot are;

The minimum value = 20

The first quartile, the 25th percentile point, Q₁ = 25

The median or second quartile, the 50th percentile point, Q₂ = 30

The third quartile, the 75th percentile point, Q₃ = 40

The maximum value = 55

The range = The maximum value - The minimum value

∴ The range = 55 - 20 = 35

We note that Nicholas earned $25 or less from the minimum value up to first quartile, Q₁, which is the 25th percentile point, where we have 25 percent of the data points

Therefore, the percent of the time Nicholas earned $25 or less is 25 percent of the time.

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n = 0.26 x 300 = 78 students.

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2 years ago
Graph the relation {(5, 0), (0, 5), (5, 1), (1, 5)}. Is it a function? Why or why not?
MissTica
D. is the correct answer, as functions cannot have the same x value correspond to different y values. For a function to be considered a function, it must always pass the vertical line test.
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4 years ago
The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra
Snezhnost [94]
Let X denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by X_1,\ldots,X_{36}, each independently and identically distributed with distribution X_i\sim\mathcal N(26,7.2).

You want to find

\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)

Recall that if X\sim\mathcal N(\mu,\sigma), then the sampling distribution \overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right) with n being the size of the sample.

Transforming to the standard normal distribution, you have

Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}

so that in this case,

Z=6\dfrac{\overline X-26}{7.2}

and the probability is equivalent to

\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)
=\mathbb P(Z>1.481)\approx0.0693
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3 years ago
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astra-53 [7]
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3 years ago
Select the number(s) that make the inequality true.<br> m &gt; 0
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Answer:

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Step-by-step explanation:

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