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Tju [1.3M]
4 years ago
10

In a study about the relationship between marital status and job level, 8,235 males at a large manufacturing firm reported their

job grade (from 1 to 4, with 4 being the highest) and their marital status. This two-way table shows the results:
Job Grade Single Married Divorced Widowed Total
1 58 874 15 8 955
2 222 3,927 10 20 4,239
3 50 2,396 34 10 2,490
4 7 533 7 4 551
Total 337 7,730 126 42 8,235

What's the conditional relative frequency among widowed employees of those widowed working at grade 2 or below?

A. 28/42 = 66.67%
B. 34/42 = 80.95%
C. 20/42 = 47.62%
D. 42/4239 = 0.1%
Mathematics
1 answer:
Katena32 [7]4 years ago
3 0

Answer:A)The conditional relative frequency among widowed employees of those widowed working at grade 2 or below= 28/42 = 66.67%


Step-by-step explanation:

In a  two way frequency table, conditional relative frequency is it’s a fraction that tells us how many elements of of a group have a certain characteristic.

Here In a study about the relationship between marital status and job level, 8,235 males at a large manufacturing firm reported their job grade (from 1 to 4, with 4 being the highest) and their marital status.

from the given two way frequency table,

Number of widowed employees at grade 1 =8

Number of widowed employees at grade 2=20

∴ Number of widowed employees at grade 2 or below=8+20=28

And the total widowed employees working =42

Now the conditional relative frequency among widowed employees of those widowed working at grade 2 or below=\frac{\text{Number of widowed employees at grade 2 or below}}{\text{Total widowed employees}}=\frac{28}{42}=0.6667=66.67%

Therefore, the conditional relative frequency among widowed employees of those widowed working at grade 2 or below= 28/42 = 66.67%

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An art teacher needs to cut out squares of paper for an art project from a left-over sheet of paper measuring 72 x 104 inches. W
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The arithmetic sequence a_ia i ​ a, start subscript, i, end subscript is defined by the formula: a_1 = 15a 1 ​ =15a, start subsc
daser333 [38]

Answer:

-1,512,390

Step-by-step explanation:

Given

a1 = 15

a_i = a_{i-1} -7

Let us generate the first three terms of the sequence

a_2 = a_{2-1}-7\\a_2 = a_1 - 7\\a_2 = 15-7\\a_2 = 8

For a_3

a_3 = a_{3-1}-7\\a_3 = a_2 - 7\\a_3 = 8-7\\a_3 = 1

Hence the first three terms ae 15, 8, 1...

This sequence forms an arithmetic progression with;

first term a = 15

common difference d = 8 - 15 = - -8 = -7

n is the number of terms = 660 (since we are looking for the sum of the first 660 terms)

Using the formula;

S_n = \frac{n}{2}[2a + (n-1)d]\\

Substitute the given values;

S_{660} = \frac{660}{2}[2(15) + (660-1)(-7)]\\S_{660} = 330[30 + (659)(-7)]\\S_{660} = 330[30 -4613]\\S_{660} = 330[-4583]\\S_{660} = -1,512,390

Hence the sum of the first 660 terms of the sequence is  -1,512,390

6 0
3 years ago
The ages of 11 students enrolled in an on-line macroeconomics course are given in the following steam-and-leaf display:
blsea [12.9K]

Answer:

The standard deviation of the age distribution is 6.2899 years.

Step-by-step explanation:

The formula to compute the standard deviation is:

SD=\sqrt{\frac{1}{n}\sum\limits^{n}_{i=1}{(x_{i}-\bar x)^{2}}}

The data provided is:

X = {19, 19, 21, 25, 25, 28, 29, 30, 31, 32, 40}

Compute the mean of the data as follows:

\bar x=\frac{1}{n}\sum\limits^{n}_{i=1}{x_{i}}

  =\frac{1}{11}\times [19+19+21+...+40]\\\\=\frac{299}{11}\\\\=27.182

Compute the standard deviation as follows:

SD=\sqrt{\frac{1}{n}\sum\limits^{n}_{i=1}{(x_{i}-\bar x)^{2}}}

      =\sqrt{\frac{1}{11-1}\times [(19-27.182)^{2}+(19-27.182)^{2}+...+(40-27.182)^{2}]}}\\\\=\sqrt{\frac{395.6364}{10}}\\\\=6.28996\\\\\approx 6.2899

Thus, the standard deviation of the age distribution is 6.2899 years.

7 0
3 years ago
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