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alexgriva [62]
3 years ago
5

Jordan's check at a restaurant was $60. She left the waiter a $12 tip. What percent is the tip of the total amount?

Mathematics
1 answer:
Finger [1]3 years ago
7 0
\frac{12}{60} is 0.2, so the answer is 20%
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A field has a length of 12 m and a diagonal of 13 m what is the width?
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Do Pythagorean Theorem because a rectangle divided in diagonals form right triangles
A^2+b^2=C^2
Plug in the correct numbers and solve
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So with correct solving you should get 5.
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This year Reuben watched 80 movies. He thought that 68 of them were very good. Of the movies he watched, what percentage did he
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He thought that 85% of them were very good.

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2 years ago
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he number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.02
Yakvenalex [24]

Answer:

a) 98.01%

b) 13.53\%

c) 27.06%

Step-by-step explanation:

Since a car has 10 square feet of plastic panel, the expected value (mean) for a car to have one flaw is 10*0.02 = 0.2  

If we call P(k) the probability that a car has k flaws then, as P follows a Poisson distribution with mean 0.2,

P(k)= \frac{0.2^ke^{-0.2}}{k!}

a)

In this case, we are looking for P(0)

P(0)= \frac{0.2^0e^{-0.2}}{0!}=e^{-0.2}=0.9801=98.01\%

So, the probability that a car has no flaws is 98.01%

b)

Ten cars have 100 square feet of plastic panel, so now the mean is 100*0.02 = 2 flaws every ten cars.

Now P(k) is the probability that 10 cars have k flaws and  

P(k)= \frac{2^ke^{-2}}{k!}

and  

P(0)= \frac{2^0e^{-2}}{0!}=0.1353=13.53\%

And the probability that 10 cars have no flaws is 13.53%

c)

Here, we are looking for P(1) with P defined as in b)

P(1)= \frac{2^1e^{-2}}{1!}=2e^{-2}=0.2706=27.06\%

Hence, the probability that at most one car has no flaws is 27.06%

6 0
3 years ago
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Solve.
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Answer:

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Step-by-step explanation:

I solved by substitution:

b = 3/4a  ---> plug this into the other equation

a + b = 21   is now    a + 3/4a = 21

Reduce: 7/4a = 21

a = 12

Now solve for b:

b=3/4a is now b=3/4(12)

b=9

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3 years ago
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If John has pairs of red, orange, yellow, blue and green socks, how many can he wear them in over 5 days, repetition is allowed
iren [92.7K]

Answer:

Total ways = 5×5×5×5×5

Total ways = 5^5

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Therefore, the correct option is C) 5^5

Step-by-step explanation:

John has pairs of red, orange, yellow, blue and green socks.

Which means that John has 5 different colors pairs of socks.

We are asked to find out in how many ways can he wear them over 5 days.

1st Day:

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2nd Day:

On the second day John has 5×5 ways to choose from.

(Since repetition is allowed)

3rd Day:

On the third day John has 5×5×5 ways to choose from.

4th Day:

On the fourth day John has 5×5×5×5 ways to choose from.

5th Day:

On the fifth day John has 5×5×5×5×5 ways to choose from.

Total ways = 5×5×5×5×5

Total ways = 5^5

Total ways = 3125

Therefore, the correct option is C) 5^5

7 0
3 years ago
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