Mathematics, 27.09.2019 05:30 LindaCat78

Each person who applies for an assembly job at robert’s electronics is given a mechanical aptitude test. one part of the test involves assembling a plug-in unit based on numbered instructions. a sample of the length of time it took 42 persons to assemble the unit was organized into the following frequency distribution. length of time (in minutes)number1 up to 444 up to 787 up to 101410 up to 13913 up to 16516 up to 192 what is the standard deviation (in minutes)?

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Answer from: kfaith88

The value of standard deviation is 3.894 minutes.

Step-by-step explanation:

Consider the provided table:

Complete the table as shown:

Length of time f class mid pt(x) f·x x² f·x²

1-4 4 2.5 10 6.25 25

4-7 8 5.5 44 30.25 242

7-10 14 8.5 119 72.25 1011.5

10-13 9 11.5 103.5 132.25 1190.25

13-16 5 14.5 72.5 210.25 1051.25

16-19 2 17.5 35 306.25 612.5

n=42 ∑f·x=384 ∑f·x²=4132.5

Now use the formula to calculate standard deviation.

$s=\sqrt{\frac{n[\sum(f\cdot x^2)]-[\sum(f\cdot x)]^2}{n(n-1)}}$

Substitute the respective values in the above formula and solve for s.

$s=\sqrt{\frac{42(4132.5)-(384)^2}{42(41)}}$

$s=\sqrt{\frac{173565-147456}{1722}}\\s=\sqrt{\frac{26109}{1722}}\\s=\sqrt{15.16}\\s\approx3.894$

Hence, the value of standard deviation is 3.894 minutes.

Answer from: Quest

-2/

What is the slope of the line? 1) -5/2 2) -2/5 3) 2/5 4) 5/2

Answer from: Quest

(a) common ratio in the given sequence is, 3

(b) $6.973568802 \times 10^9$ push ups will do kendall do on the 20th week

step-by-step explanation:

geometric sequence is a sequence of numbers that follows a pattern were the next term is found by multiplying by a constant called the common ratio(r)

(a)

given the sequence: 6 , 18 , 54 , 162 ,

this sequence is geometric sequence where each term is 3 times the previous term.

therefore,

common ratio(r) = 3.

(b)

a formula for nth term of geometric sequence is given by;

$a_n = a_1r^{n-1}$ , where r is the common ratio between the successive terms and $a_1$ is the first term,

to find how many push ups will kendall do on the 20th week.

here. n = 20th term.

in the given sequence first term $(a_1) = 6$

then;

$a_n = a_ 1 \cdot r^{n-1}$

$a_{20} = 6 \cdot 3^{20-1}$

simplify:

$a_{20} = 6.973568802 \times 10^9$

therefore, $6.973568802 \times 10^9$ push ups will do kendall do on the 20th week

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