Answer: No they don't because there is not way 1/3 can be multiplied by a number to get 2/9 and there is no way 2/9 can be divide by any number to get 1/3.

Step-by-step explanation:

4 0

3 years ago

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A school ends the year with 651 students. This is a 5% increase from the start of the school year.

Jet001 [13]

ANSWER:

n = 1.05n divided by 105 times 100

OR

n = (the number of students at the END of the school year) divided by 105 times 100

WORKING OUT:

1.05n = 651 students
1n = 651/105 times 100 = ANSWER

5 0

2 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:

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

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