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OLEGan [10]
4 years ago
5

Subtract 8 1\5 - 4 2\5 . Simplify the answer and write as a mixed number.

Mathematics
2 answers:
chubhunter [2.5K]4 years ago
4 0

For this case we must find the value of the following expression:

8 \frac {1} {5} -4 \frac {2} {5}

We have to:

8 \frac {1} {5} = \frac {5 * 8 + 1} {5} = \frac {40 + 1} {5} = \frac {41} {5}\\4 \frac {2} {5} = \frac {5 * 4 + 2} {5} = \frac {20 + 2} {5} = \frac {22} {5}

Subtracting we have:

\frac {41} {5} - \frac {22} {5} = \frac {19} {5}

In mixed number it is equivalent to:

3 \frac {4} {5}

ANswer:

3 \frac {4} {5}

kari74 [83]4 years ago
3 0

8 1\5 - 4 2\5

Make the fractions into improper fractions

8 1/5= 41/5 . Mutiply the whole number with the denominator. 8*5= 40. Add 40 with the numerator. 40+1=41

4 2/5=22/5

41/5-22/5= 19/5=3 4/5

Answer: 3 4/5

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Answer:

1) P(X = 8) = 0.1033

P(X = 9) = 0.0688

2) Expected number of 200 restaurants in which exactly 8 customers use the drive-through: 20.66

Expected number of 200 restaurants in which exactly 9 customers use the drive-through: 13.76

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In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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e = 2.71828 is the Euler number

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Question 1. Use the Poisson distribution to calculate the probability that exactly 8 cars will use the drive-through between 12:00 midnight and 12:30 AM on a Saturday night at Wendy's. Do the same for exactly 9 cars.

Cars arrive at the Wendy's drive-through at a rate of 1 car every 5 minutes between the hours of 11:00 PM and 1:00 AM. on Saturday nights. This means that during 30 minutes, 6 cars expected to arrive. So \mu = 6.

P(X = 8)

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P(X = 8) = \frac{e^{-6}*(6)^{8}}{(8)!} = 0.1033

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P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 9) = \frac{e^{-6}*(6)^{9}}{(9)!} = 0.0688

Question 2. At how many of the 200 restaurants in the survey would you expect exactly 8 customers to use the drive-through? exactly 9 customers?

There is a 10.33 probability that 8 customers would use the drive through for each restaurant.

So of 200, the expected number is

E(X) = 200*0.1033 = 20.66

There is a 6.88 probability that 9 customers would use the drive through for each restaurant.

So of 200, the expected number is

E(X) = 200*0.0688 = 13.76

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