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julia-pushkina [17]
3 years ago
5

If Mr. Kim need to 5/8 yard of wire for a project he has a stick that is 1/8 yard long how many 1/8 in your lungs links does he

need to measure to get 5/8 of a yard long
Mathematics
1 answer:
jonny [76]3 years ago
6 0

Answer:

5 stick lengths

Step-by-step explanation:

The question is not structured well. However, the following can be deduced

Wire = \frac{5}{8}yd

Stick = \frac{1}{8}yd

Required

How many of the stick length can be gotten from the wire length

To do this, we simply divide the length of the wire by the length of the stick.

So, we have:

Number = \frac{Wire}{Stick}

Number = \frac{5/8yd}{1/8yd}

Number = \frac{5/8}{1/8}

Using a calculator, we have:

Number = 5

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Answer:

0.95

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Vincent borrowed $500 from Mitchell for six months. How much interest will Mitchell earn if he charges Vincent a simple interest
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Principal P = $500,  Rate = 3 % = 3/100 = 0.03  would assume per year, 
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Simple Interest, I = P*r*t
              
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3 years ago
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is per night (USA Today, Apr
kow [346]

Answer:

P(X

Step-by-step explanation:

Assuming a mean of $204 per night and a deviation of $55.

a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)?

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean"

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the cost per night at the hotel, and for this case we know the distribution for X is given by:

X \sim N(204,55)  

Where \mu=204 and \sigma=55

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

We are interested on this probability

P(X>225)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>225)=P(\frac{X-\mu}{\sigma}>\frac{225-\mu}{\sigma})

=P(Z>\frac{225-204}{55})=P(Z>0.382)

And we can find this probability on this way:

P(Z>0.382)=1-P(Z

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B is the answer.

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