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IgorLugansk [536]
3 years ago
14

What's the answer to this?? I've tried 10 times and I can't solve it. Please help me!!

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
4 0

Answer:

whats the question

Step-by-step explanation:

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an oil change at city auto is regularly $30. Mr. Allen had a coupon for 15% off. He wants to know the sale price of the service.
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he would pay $4.50 for it.

Step-by-step explanation:

all you do is take the $30 and times that by 15% and you get $4.50.

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What is 2 1/4 equal to
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The amount of time that people spend at Grover Hot Springs is normally distributed with a mean of 73 minutes and a standard devi
Vesnalui [34]

Answer:

(a) X\sim N(\mu = 73, \sigma = 16)

(b) 0.7910

(c) 0.0401

(d) 0.6464

Step-by-step explanation:

Let <em>X</em> = amount of time that people spend at Grover Hot Springs.

The random variable <em>X</em> is normally distributed with a mean of 73 minutes and a standard deviation of 16 minutes.

(a)

The distribution of the random variable <em>X</em> is:

X\sim N(\mu = 73, \sigma = 16)

(b)

Compute the probability that a randomly selected person at the hot springs stays longer than 60 minutes as follows:

P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-73}{16})\\=P(Z>-0.8125)\\=P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays longer than an hour is 0.7910.

(c)

Compute the probability that a randomly selected person at the hot springs stays less than 45 minutes as follows:

P(X

*Use a <em>z</em>-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays less than 45 minutes is 0.0401.

(d)

Compute the probability that a randomly person spends between 60 and 90 minutes at the hot springs as follows:

P(60

*Use a <em>z</em>-table for the probability.

Thus, the probability that a randomly person spends between 60 and 90 minutes at the hot springs is 0.6464

6 0
3 years ago
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