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3 years ago

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Mathematics

1 answer:

salantis [7]3 years ago

6 0

Answer:

C. The difference of the medians is 4 times the interquartile range.

Step-by-step explanation:

The diagram show two box plots.

<u>Mahoney's box plot:</u>

Median $=33$

Interquartile range $=34-32=2$

<u>Martin's box plot:</u>

Median $=41$

Interquartile range $=42-40=2$

The interquartile ranges are the same.

The difference of the medians $=41-33=8$

Hence, the difference of the medians is 4 times the interquartile range.

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Twelve less than a number y is 8? Please help

evablogger [386]

Answer:

The answer is y = 20

Step-by-step explanation:

To find this, we need to turn it into an equation. Then we can follow the order of operations to solve.

y - 12 = 8

y = 20

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3 years ago

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The CEO of a large corporation asks his Human Resource (HR) director to study absenteeism among its executive-level managers at

Fynjy0 [20]

Answer:

Following are the answer to this question:

Step-by-step explanation:

Given:

n = 30 is the sample size.

The mean $\bar X$ = 7.3 days.

The standard deviation = 6.2 days.

df = n-1

$= 30-1 \\ =29$

The importance level is $\alpha$ = 0.10

The table value is calculated with a function excel 2010:

$= tinv (\ probility, \ freedom \ level) \\= tinv (0.10,29) \\ =1.699127\\ = t_{al(2x-1)}= 1.699127$

The method for calculating the trust interval of 90 percent for the true population means is:

Formula:

$\bar X - t_{al 2,x-1} \frac{S}{\sqrt{n}} \leq \mu \leq \bar X+ t_{al 2,x-1} \frac{S}{\sqrt{n}}$

$=\bar X - t_{0.5, 29} \frac{6.2}{\sqrt{30}} \leq \mu \leq \bar X+ t_{0.5, 29} \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 \frac{6.2}{\sqrt{30}}\leq \mu \leq7.3 +1.699127 \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 (1.13196)\leq \mu \leq7.3 +1.699127 (1.13196) \\\\=5.37 \leq \mu \leq 9.22 \\$