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alina1380 [7]
3 years ago
7

A student answered this story problem. Use estimation to check the student's answer. Lizzie has $49. Her older sister has $184.

How much money do they have altogether? Student's answer: They have a total of $233. Which estimate is correct and was the student's answer reasonable?
Mathematics
1 answer:
Anna35 [415]3 years ago
6 0

Answer:No the student was not right.Neither of the estimations were right because she did not estimate at al.

Step-by-step explanation:

1.Round 49 to the nearest ten: 50.

2.Round 184 to the nearest ten:180.

3.Add 50+180 to get 230.

The student was not right because she did not estimate 49 nor 184.

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Eva payrolls for on a number cube and the number of cubes roll 24 times then what is a reasonable prediction for the number of s
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The reasonable prediction for successful rolls is 4.

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The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per
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Answer:

1.76% probability that in one hour more than 5 clients arrive

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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:

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

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e = 2.71828 is the Euler number

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The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that \mu = 2

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

P(X \leq 5) + P(X > 5) = 1

We want P(X > 5). So

P(X > 5) = 1 - P(X \leq 5)

In which

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

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

P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353

P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707

P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707

P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804

P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902

P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824

P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176

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