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Paladinen [302]
4 years ago
12

The normal time to perform a repetitive manual assembly task is 4.25 min. In addition, an irregular work element whose normal ti

me is 1.75 min must be performed every 8 cycles. Two work units are produced each cycle. The PFD allowance factor is 16%. Determine the anticipated amount of time worked per 8-hour shift that corresponds to the PFD allowance factor of 16%.
Mathematics
1 answer:
34kurt4 years ago
8 0

Answer:

76.8 minutes

Step-by-step explanation:

Every 8 cycles we have 6 regular work elements plus 1 irregular work element.  

So, the normal time for 8 cycles is  

3*4.25+1.75 = 14.5 minutes.

Since the PDF allowance factor is 16%, the standard time for 8 cycles is 14.5 increased 16%

14.5*1.16 = 16.82 minutes.

So, the anticipated amount of time worked every 14.5 minutes

is 2.32 minutes.

An 8-hour shift has 480 minutes = 33.1034*14.5, so  the anticipated amount of time per 8-hour shift is

33.1034*2.32 = 76.8 minutes

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melomori [17]


5x[$25.85-($25.85 x 0.10) ] + (5 x 2.50)=$128.83 compared to[ 5 x 27.36( $27.36 x 0.15)]+ 10.00= 126.28 is the least amount.

answer is ------> $126.28

8 0
3 years ago
Please help me with math please, they are number lines
tekilochka [14]
The ccorrect answer is C
4 0
3 years ago
If the original quantity is 35 and the new quantity is 70​, what is the percent​ increase? Use pencil and paper. Explain what op
cricket20 [7]

Answer:

100% +

Step-by-step explanation:

35 + 35 = 70

sooooo........ +100%

4 0
3 years ago
So far a total of 48 tickets have been sold for the school play for a total revenue of $220. How many student tickets have been
Advocard [28]
Let x be the number of adult tickets and y be the number of student tickets.

From the first sentence we can set up both equations:

1) x + y = 48
2) 7x + 3y = 220

Rearrange #1 to isolate x:
x = 48 - y

Sub that into #2 and solve for y to get your answer:
7(48 - y) + 3y = 220
336 - 7y + 3y = 220
-4y = -116
y = 29

29 student tickets were sold.
6 0
3 years ago
An airline experiences a no-show rate of 6%. What is the maximum number of reservations that it could accept for a flight with a
Ber [7]

Answer:

Step-by-step explanation:

Use the normal approximation to the binomial distribution

mean µ = np

standard deviation σ = √npq

Where,

n is sample size

p is probability of success.

q is probability of failure

Given that

q = 6% =0.06

Then, p = 1-q = 1-0.06 = 0.94

Therefore:

µ = pn

µ = 0.94n

Also

σ = √npq

σ = √(n)(0.94)(0.06)

σ = √(.0564n)

Using z-scores:

z = (x — µ )/σ

Using the data above

1.645 = (160 — 0.94n)/√(0.0564n)

Cross multiply

1.645√0.0564n = 160—0.94n

Square both sides

1.645²× 0.0564n = (160-0.94n)

0.153n=25600— 300.8n + 0.8836n²

0.8836n²-300.8n-0.153n +25600=0

0.8836n² — 300.953n + 25600 = 0

Using quadratic formula method.

a = 0.8836 b = -300.953 c = 25600

n = [-b±√(b²-4ac)]/2a

n = [--300.953±√((-300.953)²-4×0.8836×25600)] / (2 × 0.8836)

n = [300.953±√(92.07)]/1.7672

n = (300.953±9.6)/1.762

n = (300.953-9.6)/1.762

n = 168.22

Or

n = (300953+9.6)/1.762

n = 176.25

The maximum number of reservation is approximately 168.

6 0
4 years ago
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