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luda_lava [24]
2 years ago
8

Percent of changeoriginal:1.5new:2.5please show work​

Mathematics
1 answer:
Luden [163]2 years ago
8 0

Answer:

Percentage change=<u>2.5-1.</u><u>5</u><u> </u> ×100℅

1.5

=66.67℅

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Assume the hold time of callers to a cable company is normally distributed with a mean of 5.5 minutes and a standard deviation o
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Answer:

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, \mu = 5.5 mins

Standard deviation, \sigma = 0.4 mins

We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.

Lets find z-score on each raw score.

⇒ z_1=\frac{x_1-\mu}{\sigma}   ...raw score,x_1 = 5.4

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⇒ z_1=\frac{5.4-5.5}{0.4}

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For raw score 5.5 the z score is.

⇒ z_2=\frac{5.8-5.5}{0.4}  

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Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.

We have to work with P(5.4<z<5.8).

⇒ P(5.4

⇒ P(-0.25

⇒ P(z

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⇒ 0.7734-0.4013

⇒ 0.3721

To find the percentage we have to multiply with 100.

⇒ 0.3721\times 100

⇒ 37.21 %

The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21

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