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Gekata [30.6K]
1 year ago
9

Worker Efficiency An efficiency study conducted for Elektra Electronics showed that the number of Space Commander walkie-talkies

assembled by the average worker during the morning shift t hr after starting work at 8 a.m. is given by
N(t) = −t3 + 6t2 + 15t (0 ≤ t ≤ 4).

(a) Find the rate at which the average worker will be assembling walkie-talkies t hr after starting work.
(b) At what rate will the average worker be assembling walkie-talkies, per hour, at 9 a.m.?walkie-talkies/hour At what rate will the average worker be assembling walkie-talkies, per hour, at 10 a.m.?
(c) How many walkie-talkies will the average worker assemble between 9 a.m. and 10 a.m.?walkie-talkies
Mathematics
1 answer:
nalin [4]1 year ago
4 0

Answer:

a) -2t²+12t+15    b) 25, 31   c) 26

Step-by-step explanation:

a) To find out rate, we will take derivative of the given equation

dN/dt= -2t2+12t+15

b) At 9 am , it will be 1hr since the start of morning shift so t=1

dN/dt= -2(1)²+12(1)+15

dN/dt=25

At 10 am, it will  2 hrs since the start of morining shift be so t=2

dN/dt=-2(2)²+12(2)+15

DN/dt= 31

c) N(2)-N(1)= (-(2)³+6(2)²+15(2))-(-(1)³+6(1)²+15(1))

                =(46)-(20))

                = 26

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Answer:

The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).

Step-by-step explanation:

Intern            No. of Breast

Number        Exams Performed               X²

1                         30                                  900

2                        40                                 1600

3                        8                                        64

4                        20                                   400

5                          26                                 676

6                           35                               1225

7                            35                               1225

8                            20                                400

9                             25                              625

<u>10                                 20                        400 </u>

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Mean= X`= ∑x/n= 259/10= 25.9

Variance = s²= 1/n-1[∑X²- (∑x)²/n]

= 1/0[7515- (259)²/10]= 1/9[7515- 6708.1]

= 806.9/9=89.655= 89.66

Standard Deviation= √89.655= 9.4687

Hence

The value of t with significance level alpha= 0.05 and 9 degrees of freedom  is t(0.025,9)= 2.262

The 95 % Confidence interval is given by

x`±t(∝,n-1) s/√n

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25.9± 2.262( 9.4687/√10)

= 25.9 ±2.262 (2.9943)

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= 25.9 +6.7730=32.6730

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