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kati45 [8]
3 years ago
10

The owner of a computer repair shop has determined that their daily revenue has mean​ $7200 and standard deviation​ $1200. The d

aily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed​ $7500? Round to four decimal places. A. 0.9131 B. 0.0869 C. 0.9147 D. 0.0853
Mathematics
1 answer:
liraira [26]3 years ago
7 0

Answer: D.0.0853

Step-by-step explanation:

We assume that the daily revenue totals for the next 30 days will be follow a normal distribution.

Given : Population mean : \mu=\$7200

Standard deviation : \sigma=\​ $1200

Sample size : n=30

Let X be the random variable that represents the daily revenue .

Z-score : z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For x=$7500

z=\dfrac{7500-7200}{\dfrac{1200}{\sqrt{30}}}\approx1.37

By using the standard normal distribution table for z , we have

P(x>7500)=P(z>1.37)=1-P(z\leq1.37)

=1- 0.9146565= 0.0853435\approx0.0853

Hence, the probability that the mean daily revenue for the next 30 days will exceed​ $7500= 0.0853=

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Answer:

75.1^{\circ}

Step-by-step explanation:

In this problem, we have:

H = 452 m is the height of the Petronas tower

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d = 120 m is the distance between the woman and the base of the tower

First of all, we notice that we want to find the angle of elevation between the woman's hat the top of the tower; this means that we have consider the difference between the height of the tower and the height of the woman, so

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Now we notice that d and H' are the two sides of a right triangle, in which the angle of elevation is \theta. Therefore, we can write the following relationship:

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since

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\theta = tan^{-1}(\frac{H'}{d})=tan^{-1}(\frac{450.25}{120})=75.1^{\circ}

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3 years ago
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nikitadnepr [17]

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Step-by-step explanation:

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