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scoray [572]
3 years ago
8

Can someone please, please, please help me with my algebra questions? Specifically with statistics and probability. I have tried

all week to find someone to help me, that knows the correct answer and can provide me an explanation behind it. I've sent messages to people and left comments but no one will respond to me, and if they do, they never answer my questions. I need to have a good grade in my Algebra class with all correct answers on my assignments because I'm horrible with math in general. Please don't ask me to post my questions, when they're already on my profile. I desperately need help with my algebra problems because I'm at my breaking point on here.
Mathematics
1 answer:
Vikentia [17]3 years ago
7 0

Answer:

You can message me on here. As long as it's not Algebra II we are good. I have completed Algebra I honors and am about to complete Geometry Honors. I will try to help you out as much as I can.

Step-by-step explanation:

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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
Ierofanga [76]

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

⇒ Plugging the values.

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For raw score 5.5 the z score is.

⇒ z_2=\frac{5.8-5.5}{0.4}  

⇒ z_2=0.75

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).

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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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Step-by-step explanation:

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