What do banana and grammar have in common?

Citrus2011 [14]

Hi!

There are two ways to do this.

1. Start with the first part of the decimal, (4.) then divide 3/7 to get the second part.

3/7 = 0.43

4.43

2. Convert the mixed number 4 3/7 to a fraction.

4 x 7 = 28
28 + 3 = 31

31/7

Now convert that into a decimal.

31/7 = 4.43

The answer is 4.43

Hope this helps! :)

3 0

3 years ago

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17) If your GROSS INCOME is $90,000 and you have 33% in TAXES

pychu [463]

60,000 nnnmmmmmmmmmm

5 0

3 years ago

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]

Answer:

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:

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

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

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

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:

$\\ P(z$

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>

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

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

$\\ P(z$

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

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

$\\ z_{score}=1.745454$

$\\ P(z$

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>

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

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