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NemiM [27]
3 years ago
5

Mrs. Wright bought some notebook paper for her class. She decided to keep 3/4 of the paper for her own use. How much paper did s

he have left to share with her students?
Mathematics
1 answer:
larisa [96]3 years ago
5 0
So the total amount of paper is 4/4 right?
And she kept 3/4.
So we have to minus 3/4 from 4/4.
Because it has the same denominator we need not change it to similar one.So 4-3 is one.
So answer is 1/4
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A major traffic problem in the Greater Cincinnati area involves traffic attempting to cross the Ohio River from Cincinnati to Ke
yaroslaw [1]

Answer:

a. 0.563 = 56.3% probability that for the next 60 minutes (two time periods) the system will be in the delay state.

b. 0.625 = 62.5% probability that in the long run the traffic will not be in the delay state

Step-by-step explanation:

Question a:

The probability of finding a traffic delay in one period, given a delay in the preceding period, is 0.75.

The system currently is in traffic delay, so for the next time period, 0.75 probability of a traffic delay. If the next period is in a traffic delay, the following period will also have a 0.75 probability of a traffic delay. So

0.75*0.75 = 0.563

0.563 = 56.3% probability that for the next 60 minutes (two time periods) the system will be in the delay state.

b. What is the probability that in the long run the traffic will not be in the delay state? If required, round your answers to three decimal places.

If it doesn't have a delay, 85% probability of continuing without a delay.

If it has a delay, 75% probability of continuing with a delay.

So, for the long run:

x: current state

85% probability of no delay if x is in no delay, 100 - 75 = 25% if x is in delay(1-x). So

0.85x + 0.25(1 - x) = x

0.6x + 0.25 = x

0.4x = 0.25

x = \frac{0.25}{0.4}

x = 0.625

0.625 = 62.5% probability that in the long run the traffic will not be in the delay state

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2 years ago
OKEASE HELP HELP HELP
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Answer:

C

Step-by-step explanation:

cus irs CCCCCCcCChdjsikdkrkf

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Is this answer right??
Serhud [2]

Answer:

I think so

Step-by-step explanation:

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