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KATRIN_1 [288]
3 years ago
15

PLEASE HELP ME WITH THE CORRECT ANSWER !!

Mathematics
2 answers:
Harrizon [31]3 years ago
7 0
30 im positive is the answer
leva [86]3 years ago
7 0
The answer is 30 kg if each one ways 6 kg multiply 6 and 5.
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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.

⇒ z_1=\frac{5.4-5.5}{0.4}

⇒ z_1=-0.25  

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).

⇒ P(5.4

⇒ P(-0.25

⇒ P(z

⇒ z(1.5)=0.7734 and z(-0.25)=0.4013.<em>..from z -score table.</em>

⇒ 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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3 years ago
.Write in standard form. 4h - 6(5+7h)
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=−38h+−30

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1

Step-by-step explanation:

2^−1 + 2^−1

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