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kirza4 [7]
3 years ago
14

How would you estimate whether 27/50 is closer to 1/2 or 1 without using a number line? Explain.

Mathematics
1 answer:
Ivahew [28]3 years ago
7 0

Answer

Find\ out\ the\ how\ would\ you\ estimate\ whether\ \frac{27}{50}\ is\ closer\ \frac{1}{2}  to\ or 1\ without\ using\ a\ number\ line .

To prove

As\ the\ value\ of\ \frac{27}{50}\ in\ decimal\ form.

= \frac{27}{50}

= 0.54

As\ the\ value\ of\ \frac{1}{2}\ in\ decimal\ form.

= \frac{1}{2}

= 0.5

As\ the\ value\ of\ 1 \ in\ decimal\ form.

= 1.00

Now we find out the mid point of the 0.5 and 1.00 .

Mid\ point = \frac{1.00+0.5}{2}

(Mid point is the point lies middle of a line segment .)

Mid point = 0.75

Thus

0.54< 0.75

This shows o.54 lies between 0.5 and 1.00

This means 0.54 lies right to 0.5 and left from 1.00 .

Therefore

\frac{27}{50}\ lies\ right\ to\ \frac{1}{2} and\ left\ from\ 1.



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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 Saturd
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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

Step-by-step explanation:

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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In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

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)

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

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

P(X = 9)

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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