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vazorg [7]
3 years ago
15

Which expression does not represent "the sum of n and 6"? n + 6 6 n 6 + n

Mathematics
2 answers:
olga2289 [7]3 years ago
8 0

Answer:

Step-by-step explanation:

6n because 6n it's 6 times n

Misha Larkins [42]3 years ago
4 0

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<em>Hey</em><em>!</em><em>!</em>

<em> </em><em>Answer</em><em>:</em><em> </em><em>6</em><em>n</em>

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<em>6</em><em>n</em><em> </em><em>represe</em><em>nts</em><em> </em><em>"</em><em>the</em><em> </em><em>product</em><em> </em><em>of</em><em> </em><em>n</em><em> </em><em>and</em><em> </em><em>6</em><em>"</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

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<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

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Step-by-step explanation:

you first do a vertical cut

then a horizontal cut

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The length of a metal rod is 9 3/4 feet. What is the length of 1/4 of the rod?
7nadin3 [17]

Answer:

2 \frac{5}{16}

Step-by-step explanation:

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Simplify 2(x - 3) + 7(x + 2).<br><br> 9x - 8<br> 9x + 3<br> 9x + 8
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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
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

8 0
3 years ago
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