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algol13
3 years ago
5

Tell what whole number you can substitute for A in each list so the numbers are ordered from least to greatest. If there is none

, explain why. 3/A, A/5, 75% (P.S those are supposed to be fractions, and A is an algebra thingy)
Mathematics
1 answer:
hodyreva [135]3 years ago
4 0

Answer: None

Step-by-step explanation:

If A=4, say for example, our list becomes: 3/4, 4/5, 75%.

This in decimal form becomes: 0.75, 0.8, 0.75 which is clearly not in ascending order.

If A =5, the list becomes 3/5, 5/5, 75% which clearly is 0.6, 1, 0.75 which is also not in ascending order.

For any value of A greater than 5, the second term A/5 is greater than 1 thus being less than the third term 0.75.

Therefore there is no whole number A that makes the list true.

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$1200/ 100 square feet         multiply 11 on both sides

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<h3><u><em>Please mark this as brinliest thank you!!!</em></u></h3><h3><u><em>: )</em></u></h3><h3><u><em>And as always, </em></u></h3><h3><u><em>simplify bananas</em></u></h3>
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Step-by-step explanation:

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ICE Princess25 [194]

Answer:

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

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:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

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

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

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