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3 years ago

Simplify the following questions: 2[y-3(y-5)]

2 answers:

7 0

2(y−3(y−5))=(2)(y+−3y+15)=(2)(y)+(2)(−3y+15)=2y−6y+30=−4y+30

Sauron [17]3 years ago

4 0

The answer would be -4y+30

2[y-3(y-5)]

2y-6(y-5)

2y-6y-30

4y-30

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Finn changes his mind and, from now on, decides to take the normal route to work everyday. On any given day, the time (in minute

Margaret [11]

Answer:

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

Step-by-step explanation:

This can be solved by the the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

Each z-score value has an equivalent p-value, that represents the percentile that the value X is:

The problem states that:

Mean = 35, so $\mu = 35$

Variance = 81. The standard deviation is the square root of the variance, so $\sigma = \sqrt{81} = 9$ .

Find the 33rd percentile of the time it takes Finn to get to work on any given day. Do not include any units in your answer.

Looking at the z-score table, $z = -0.44$ has a pvalue of 0.333. So what is the value of X when $z = -0.44$ .

$Z = \frac{X - \mu}{\sigma}$

$-0.44 = \frac{X - 35}{9}$

$X - 35 = -3.96$

$X = 31.04$

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

Over the next 2 days, find the probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that Finn took more than 40.5 minutes to get to work on the first day. The first step to solve this problem is finding the z-value of $X = 40.5$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40.5 - 35}{9}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.7291. This means that the probability that it took LESS than 40.5 minutes for Finn to get to work is 72.91%. The probability that it took more than 40.5 minutes if $P_{1} = 100% - 72.91% = 27.09% = 0.2709$

$P_{2}$ is the probability that Finn took more than 38.5 minutes to get to work on the second day. Sine the probabilities are independent, we can solve it the same way we did for the first day, we find the z-score of

$X = 38.5$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{38.5 - 35}{9}$

$Z = 0.39$

$Z = 0.39$ has a pvalue of 0.6517. This means that the probability that it took LESS than 38.5 minutes for Finn to get to work is 65.17%. The probability that it took more than 38 minutes if $P_{1} = 100% - 65.17% = 34.83% = 0.3483$

So:

$P = P_{1} + P_{2} = 0.2709 + 0.3483 = 0.6192$

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

7 0

3 years ago

You put $500 in a savings account. The account earns $15.75 simple interest in 6 months. What is the annual interest rate? Show

marta [7]

So in 6 months (1/2yr) you earned $15.75 interest, if we double that we find the total interest you would gain for that year, 15.75x2= $31.5 total interest for that year.

To find the annual rate divide the interest gained by the amount deposited, 31.5/500 = 0.063% p.a.

3 0

3 years ago

The 10. A jet flies at an altitude of 52,800 feet. What is the height of the jet in miles?

aliya0001 [1]

Answer:

The Answer is 10 Miles.

Step-by-step explanation:

The jet is 52,800 feet high.

There are 5,280 feet in a mile.

Divide 52,800 by 5280 to find miles:

52800 / 5280 = 10 Miles.

Hope this Helps! Have an Awesome Day!! :-)

5 0

3 years ago

Read 2 more answers

Can somebody please help me out

Alekssandra [29.7K]

Answer:

with what????

8 0

3 years ago

How do I find the answer to m

ZanzabumX [31]

Those are vertical angles,so they are equal
150=5m
m=150/5
m=30

3 0

3 years ago