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3 years ago

What is the sum of -7-5+3

Mathematics

2 answers:

AysviL [449]3 years ago

8 0

The answer to your question is -9

pentagon [3]3 years ago

7 0

The answer would be -9

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An ice cream truck was parked near a beach, and some kids ran to it and formed a line. Then, more kids joined the line by fillin

Irina-Kira [14]

Answer:

Step-by-step explanation:

So, we know that this process happened 3 times in total. The original amount of kids we can call x, so the next wave filled in the gaps inbetween, so we call that amount x-1, the next wave would be (x+x-1)-1 or 2x-2, and the last would be 4x-4

adding this all up we get 8x-7=49, so 8x = 56 or x = 7

3 0

2 years ago

What is the answer to 23×-5

Ray Of Light [21]

Answer:

23×-5

= -115

The answer is -115, easy peasy

5 0

3 years ago

Suppose that on the average, 7 students enrolled in a small liberal arts college have their automobiles stolen during the semest

sineoko [7]

Answer:

0.91824 = 91.824% probability that more than 3 students will have their automobiles stolen during the current semester.

Step-by-step explanation:

We have only the mean, which means that the Poisson distribution is used to solve this question.

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 interval.

Suppose that on the average, 7 students enrolled in a small liberal arts college have their automobiles stolen during the semester.

This means that $\mu = 7$

What is the probability that more than 3 students will have their automobiles stolen during the current semester?

This is:

$P(X > 3) = 1 - P(X \leq 3)$

In which

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

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

$P(X = 0) = \frac{e^{-7}*7^{0}}{(0)!} = 0.00091$

$P(X = 1) = \frac{e^{-7}*7^{1}}{(1)!} = 0.00638$

$P(X = 2) = \frac{e^{-7}*7^{2}}{(2)!} = 0.02234$

$P(X = 3) = \frac{e^{-7}*7^{3}}{(3)!} = 0.05213$

Then

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.00091 + 0.00638 + 0.02234 + 0.05213 = 0.08176
$

$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.08176 = 0.91824$

0.91824 = 91.824% probability that more than 3 students will have their automobiles stolen during the current semester.

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Ilia_Sergeevich [38]

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