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salantis [7]
3 years ago
6

In a science class student made rockets out of empty 2-liter bottles. When launched, Karen's rocket reached a height of 12 and 1

/2 ft. Sam's rocket reached a height 1 and 2/5 times greater than Karen's rocket. How high did sam's rocket go?
Mathematics
2 answers:
alekssr [168]3 years ago
5 0
Just add 12 1/2 to 1 2/5 and remember that the bottoms of fractions have to be the same and if you change the bottoms, you have to change the top just as much :)13 9/10 is the answer
NikAS [45]3 years ago
4 0
The answer is 24 2/10. 12 1/2 x 2/5.
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At the beginning of the month Leslie's parents put $35.00 on her account. If lunch costs $2.55 per day, how much money would be
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$73.25

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New York City is the most expensive city in the United States for lodging. The room rate is $204 per night (USA Today, April 30,
Sever21 [200]

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a. 0.35197 or 35.20%; b. 0.1230 or 12.30%; c. 0.48784 or 48.78%; d. $250.20 or more.

Step-by-step explanation:

In general, we can solve this question using the <em>standard normal distribution</em>, whose values are valid for any <em>normally distributed data</em>, provided that they are previously transformed to <em>z-scores</em>. After having these z-scores, we can consult the table to finally obtain the probability associated with that value. Likewise, for a given probability, we can find, using the same table, the z-score associated to solve the value <em>x</em> of the equation for the formula of z-scores.

We know that the room rates are <em>normally distributed</em> with a <em>population mean</em> and a <em>population standard deviation</em> of (according to the cited source in the question):

\\ \mu = \$204 <em>(population mean)</em>

\\ \sigma = \$55 <em>(population standard deviation)</em>

A <em>z-score</em> is the needed value to consult the <em>standard normal table. </em>It is a transformation of the data so that we can consult this standard normal table to obtain the probabilities associated. The standard normal table has a mean  of 0 and a standard deviation of 1.

\\ z_{score}=\frac{x-\mu}{\sigma}

After having all this information, we can proceed as follows:

<h3>What is the probability that a hotel room costs $225 or more per night? </h3>

1. We need to calculate the z-score associated with x = $225.

\\ z_{score}=\frac{225-204}{55}

\\ z_{score}=0.381818

\\ z_{score}=0.38

We rounded the value to two decimals since the <em>cumulative standard normal table </em>(values for cumulative probabilities from negative infinity to the value x) to consult only have until two decimals for z values.

Then

2. For a z = 0.38, the corresponding probability is P(z<0.38) = 0.64803. But the question is asking for values greater than this value, then:

\\ P(z>038) = 1 - P(z (that is, the complement of the area)

\\ P(z>038) = 1 - 0.64803

\\ P(z>038) = 0.35197

So, the probability that a hotel room costs $225 or more per night is P(x>$225) = 0.35197 or 35.20%, approximately.

<h3>What is the probability that a hotel room costs less than $140 per night?</h3>

We follow a similar procedure as before, so:

\\ z_{score}=\frac{x-\mu}{\sigma}

\\ z_{score}=\frac{140-204}{55}

\\ z_{score}=\frac{140-204}{55}

\\ z_{score}= -1.163636 \approx -1.16

This value is below the mean (it has a negative sign). The standard normal tables does not have these values. However, we can find them subtracting the value of the probability obtained for z = 1.16 from 1, since the symmetry for normal distribution permits it. Then, the probability associated with z = -1.16 is:

\\ P(z

\\ P(z

\\ P(z

Then, the probability that a hotel room costs less than $140 per night is P(x<$140) = 0.1230 or 12.30%.

<h3>What is the probability that a hotel room costs between $200 and $300 per night?</h3>

\\ z_{score}=\frac{x-\mu}{\sigma}

<em>The z-score and probability for x = $200:</em>

\\ z_{score}=\frac{200-204}{55}

\\ z_{score}= -0.072727 \approx -0.07

\\ P(z

\\ P(z

\\ P(z

<em>The z-score and probability for x = $300:</em>

\\ z_{score}=\frac{300-204}{55}

\\ z_{score}=1.745454

\\ P(z

\\ P(z

\\ P(z

Then, the probability that a hotel room costs between $200 and $300 per night is 0.48784 or 48.78%.

<h3>What is the cost of the most expensive 20% of hotel rooms in New York City?</h3>

A way to solve this is as follows: we need to consult, using the cumulative standard normal table, the value for z such as the probability is 80%. This value is, approximately, z = 0.84. Then, solving the next equation for <em>x:</em>

\\ z_{score}=\frac{x-\mu}{\sigma}

\\ 0.84=\frac{x-204}{55}

\\ 0.84*55=x-204

\\ 0.84*55 + 204 =x

\\ x = 250.2

That is, the cost of the most expensive 20% of hotel rooms in New York City are of $250.20 or more.

6 0
3 years ago
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