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antoniya [11.8K]
2 years ago
10

A community college has 150 word processors. The probability that any one of them will require repair on a given day is 0.025. T

o find the probability that exactly 25 of the word processors will require repair, one will use what type of probability distribution?
Mathematics
1 answer:
Volgvan2 years ago
6 0

Answer:

Binomial Distribution

Step-by-step explanation:

In this case there are only two possible outcomes that is either the processor requires repair or does not require.

In such cases binomial distribution is beneficial to use.

Binomial Distribution is simply the probability of failure or success in an experiment that is repeated multiple times.

for binomial to be used following three conditions are to be used:

1. Fixed number of trails

2. Each trial or observation is independent

3. probability of success is exactly same

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natali 33 [55]
X = 112
See attachment file below.

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5 0
2 years ago
Which translation maps the graph of the function f(x) = x2 onto the function g(x) = x2 + 2x + 6?
Flura [38]
The graph went left 1 and up 6. You could complete the square to find out or just graph it!
Hope this helps!
4 0
2 years ago
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The equation y = 5.5x represents a proportional relationship. What is the constant of proportionality?
lara31 [8.8K]
Y = 5.5x
The constant is 5.5 
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5 0
2 years ago
Jane invested £4000 for 3 years with an interest rate of 1.5%. What was her investment worth at the end of this period?
ddd [48]

Answer:

$4182.7

Step-by-step explanation:= 4000*(1.5%)*3

Year 1= 4000*(100%+1.5%)= 4060

Year 2= 4060*(100%+1.5%)= 4120.9

Year 3= 4120.9*(100%+1.5%)= 4182.7

6 0
2 years ago
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A company manufacturing oil seals wants to establish X and R control charts on the process. There are 25 preliminary samples of
alisha [4.7K]

Answer:

A ) i) X control chart : upper limit = 50.475, lower limit = 49.825

    ii) R control chart : upper limit =  1.191, lower limit = 0

Step-by-step explanation:

A) Finding the control limits

grand sample mean = 1253.75 / 25 = 50.15

mean range = 14.08 / 25 = 0.5632

Based on  X control CHART

The upper control limit ( UCL ) =

grand sample mean + A2* mean range ) = 50.15 + 0.577(0.5632) = 50.475

The lower control limit (LCL)=

grand sample mean - A2 *  mean range = 50.15 - 0.577(0.5632) = 49.825

Based on  R control charts

The upper limit = D4 * mean range = 2.114 * 0.5632 = 1.191

The lower control limit = D3 * mean range = 0 * 0.5632 = 0  

B) estimate the process mean and standard deviation

estimated process mean = 50.15 = grand sample mean

standard deviation = mean range / d2  = 0.5632 / 2.326 = 0.2421

note d2 is obtained from control table

6 0
2 years ago
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