Mathematics : asked on sksksjs
 30.08.2022

Evaluate the equation. f + 14 2/7= 23

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24.06.2023, solved by verified expert
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Third item in the list

Step-by-step explanation:

Third function:  f(x) = x + 9.  If we substitute 14 for x, we get f(14) = 14 + 9 = 23.

Step-by-step explanation:

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Mathematics
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P Answered by PhD

C

Step-by-step explanation:

f+98/7=23

f=161/7-98/7=63/7=9

C

Mathematics
Step-by-step answer
P Answered by PhD

Third item in the list

Step-by-step explanation:

Third function:  f(x) = x + 9.  If we substitute 14 for x, we get f(14) = 14 + 9 = 23.

Step-by-step explanation:

Mathematics
Step-by-step answer
P Answered by Specialist

The null hypothesis is that there is no difference between the mean time it takes an online trained employee or a team-based trained employee to assemble the given part

Step-by-step explanation:

The information the director of the equipment manufacturing company is interested in determining is weather the productivity of assembly line employees is affected by the method used in their training

The total number newly hired employees in the sample = 42

The number of newly hired employees that receive training online = 21

The number of newly hired employees that receive training in a team = 21

The given data of the result of the time it takes an employee to assemble a part is presented as follows;

On-Line 19.4, 16.7, 20.7, 19.3, 21.8, 16.8, 14.1, 17.7, 16.1, 19.8, 16.8, 19.3, 14.7, 16.0, 16.5, 17.7, 16.2, 17.4, 16.4, 16.8, 18.5

Team; 22.4, 13.8, 18.7, 18.0, 19.3, 20.8, 15.6, 17.1, 18.0, 28.2, 21.7, 20.8, 30.7, 24.7, 23.7, 17.4, 23.2, 20.1, 12.3, 15.2, 16.0

The mean time for the of the on-line trained employee, \overline x_1 = 17.55714

The standard deviation of the time for the of the on-line trained employee, s₁ = 1.93328

The mean time for the of the team based trained employee, \overline x_2 = 19.89048

The standard deviation of the time for the of the team based trained employee, s₂ = 4.5667

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1\overline x_2

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}

Therefore, we have;

t=\dfrac{(19.89048 - 17.55714)}{\sqrt{\dfrac{4.57667^{2}}{21} - \dfrac{1.93328^{2} }{21}}} \approx 2.5776

The degrees of freedom, df = n₁ - 1 = 21 - 1 = 20

At 95% confidence level, we have α = 1 - 0.95 = 0.05, and t = 2.086

Therefore, given that the test statistic is larger than the critical 't' value, we reject the null hypothesis. There is sufficient statistical evidence to show that there is a difference between the mean time of assembly value for the on-line trained and team-based employee

Mathematics
Step-by-step answer
P Answered by Master

The null hypothesis is that there is no difference between the mean time it takes an online trained employee or a team-based trained employee to assemble the given part

Step-by-step explanation:

The information the director of the equipment manufacturing company is interested in determining is weather the productivity of assembly line employees is affected by the method used in their training

The total number newly hired employees in the sample = 42

The number of newly hired employees that receive training online = 21

The number of newly hired employees that receive training in a team = 21

The given data of the result of the time it takes an employee to assemble a part is presented as follows;

On-Line 19.4, 16.7, 20.7, 19.3, 21.8, 16.8, 14.1, 17.7, 16.1, 19.8, 16.8, 19.3, 14.7, 16.0, 16.5, 17.7, 16.2, 17.4, 16.4, 16.8, 18.5

Team; 22.4, 13.8, 18.7, 18.0, 19.3, 20.8, 15.6, 17.1, 18.0, 28.2, 21.7, 20.8, 30.7, 24.7, 23.7, 17.4, 23.2, 20.1, 12.3, 15.2, 16.0

The mean time for the of the on-line trained employee, \overline x_1 = 17.55714

The standard deviation of the time for the of the on-line trained employee, s₁ = 1.93328

The mean time for the of the team based trained employee, \overline x_2 = 19.89048

The standard deviation of the time for the of the team based trained employee, s₂ = 4.5667

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1\overline x_2

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}

Therefore, we have;

t=\dfrac{(19.89048 - 17.55714)}{\sqrt{\dfrac{4.57667^{2}}{21} - \dfrac{1.93328^{2} }{21}}} \approx 2.5776

The degrees of freedom, df = n₁ - 1 = 21 - 1 = 20

At 95% confidence level, we have α = 1 - 0.95 = 0.05, and t = 2.086

Therefore, given that the test statistic is larger than the critical 't' value, we reject the null hypothesis. There is sufficient statistical evidence to show that there is a difference between the mean time of assembly value for the on-line trained and team-based employee

Mathematics
Step-by-step answer
P Answered by PhD

Step-by-step explanation:

(1)  3 1/4 = 13 / 4

     (13 / 4) / 2 = 13 / 8 = 1 5/8

(2) (1 / 7) / (2 / 5) = (1 * 5) / ( 2 * 7) = 5 / 14

5 / 14 - 1 / 7 = 5 / 14 - 2 / 14 = 3 / 14

(3) 6 1/2 = 13 / 2    and 3 2/3 = 11/3

(13 /2) / (11 / 3) = (3 * 13) / (2 * 11) = 39 / 22 = = 1 17/22

(4) 3 1/5 = 16 / 5    

16 / 5 * 5 / 8 = 80 / 40 = 2

(5) 2 3/4 = 11 / 4      

the reciprocal is 4 / 11

Mathematics
Step-by-step answer
P Answered by PhD

Step-by-step explanation:

(1)  3 1/4 = 13 / 4

     (13 / 4) / 2 = 13 / 8 = 1 5/8

(2) (1 / 7) / (2 / 5) = (1 * 5) / ( 2 * 7) = 5 / 14

5 / 14 - 1 / 7 = 5 / 14 - 2 / 14 = 3 / 14

(3) 6 1/2 = 13 / 2    and 3 2/3 = 11/3

(13 /2) / (11 / 3) = (3 * 13) / (2 * 11) = 39 / 22 = = 1 17/22

(4) 3 1/5 = 16 / 5    

16 / 5 * 5 / 8 = 80 / 40 = 2

(5) 2 3/4 = 11 / 4      

the reciprocal is 4 / 11

Business
Step-by-step answer
P Answered by PhD

B. The total hiring cost is $62,400

D. The total hiring cost is larger than the total firing cost.

E. Overtime labor cost is $200,000.

Explanation:

The hiring cost of the labor is greater than the firing cost because the company needs more workforce to meet the finished goods demand in the certain period. The hiring cost of the employees is $800 per labor and company needs 78 more labor to meet the demand. The total hiring cost will be $800 * 78 labors = $62,400.

Business
Step-by-step answer
P Answered by PhD

B. The total hiring cost is $62,400

D. The total hiring cost is larger than the total firing cost.

E. Overtime labor cost is $200,000.

Explanation:

The hiring cost of the labor is greater than the firing cost because the company needs more workforce to meet the finished goods demand in the certain period. The hiring cost of the employees is $800 per labor and company needs 78 more labor to meet the demand. The total hiring cost will be $800 * 78 labors = $62,400.

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