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2 years ago

Need help with this question

Mathematics

2 answers:

Dmitrij [34]2 years ago

4 0

Answer:

A) -9

B)-9+-9= -18

C) -8

D) 0

gtnhenbr [62]2 years ago

4 0

The answer is

-9+9

Because it’s a -9 and you’re adding 9 it is 0.

So the answer is -9+9

MARK ME BRAINLIEST PLEASE!!!!!!

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The number of typing errors made by a typist has a Poisson distribution with an average of three errors per page. If more than t

damaskus [11]

Answer:

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

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 interval.

Poisson distribution with an average of three errors per page

This means that $\mu = 3$

What is the probability that a randomly selected page does not need to be retyped?

Probability of at most 3 errors, so:

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

In which

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

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

Then

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472$

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

3 0

2 years ago

How do you add integers with different sgns

Anuta_ua [19.1K]

Adding integers with different signs is just like adding or subtracting. First you just have to add all negative integers and also add all positive. After that positive intergers will be deducted by the sum of negative integers

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3 years ago

PLZ HELP WILL GIVE BRAINLIEST WORTH 10 PTS THX!!

antiseptic1488 [7]

Answer:

64/49

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)^c = a^(bc)

(a^b)(a^c) = a^(b+c)

a^-b = 1/a^b

a^0 = 1

_____

Using these rules, we can simplify the given expression to ...

$\left(\dfrac{8\cdot 4\cdot 2}{8\cdot 7}\right)^2\times\left(\dfrac{8^0}{7^{-3}}\right)^3\times 7^{-9}\\\\=\dfrac{8^2}{7^2}\times(8^{0\cdot 3}\cdot 7^{3\cdot 3})\times 7^{-9}\\\\=\dfrac{64}{49}\times(8^07^{9-9})=\boxed{\dfrac{64}{49}}$