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Savatey [412]
3 years ago
12

Greg assembles units of a certain product at a factory. Each day he is paid $2.00 per unit for the first 40 units that he assemb

les and $2.50 for each additional unit that he assembles that day. If Greg assembled at least 30 units on each of two days and was paid a total of $180.00 for assembling units on the two days, what is the greatest possible number of units that he could have assembled on one of the two days
Mathematics
1 answer:
levacccp [35]3 years ago
7 0

Answer:

45 on each day at least thats what i think

Step-by-step explanation:

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Answer:

Step-by-step explanation:

To calculate ;

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= 1 x 0.05 + 2 x 0.09 + 3 x 0.03 + 4 x 0.42 + 5 x 0.41

= 4.05

2) the expected value of the job satisfaction score for middle managers;

= 1 x 0.04 + 2 x 0.10 + 3 x 0.12 + 4 x 0.46 + 5 x 0.28

= 3.84

c) the variance of job satisfaction scores for executives and middle managers (to 2 decimals).

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ii) for middle managers ; 1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 = 1.134 = 1.13

d) the standard deviation of job satisfaction scores for both probability distributions (to 2 decimals). Executives, Middle managers;

For Executives = square root [ 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2] = 1.12

For Middle Managers ; Square root [1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 ] = 1.06

e) from the values gotten for the variance of both executive and middle managers, the variance of the former is more than that of the latter as such higher satisfaction with the executive managers.

5 0
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