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2 years ago

Whats 2(x+4) Plss help

Mathematics

2 answers:

abruzzese [7]2 years ago

7 0

Answer:

2x+8

Step-by-step explanation:

2 times x = 2x. 2 times 4 = 8.

emmainna [20.7K]2 years ago

6 0

Answer:

2x+8

Step-by-step explanation:

You multiply (x+4) by 2. So 2(x) = 2 and 2(4) = 8. So 2x+8.

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HURRY I NEED THIS BEFORE 12

I am Lyosha [343]

X=8 because add 11 to both sides and it becomes 28=3.5x and divide both sides by 3.5 so x=8

5 0

2 years ago

If <br> B is the midpoint of AC, AB=x+6 and AC=5x-6, then what is BC?

nikklg [1K]

If B is the midpoint of AC, then AB=BC. It is also BC=AC-AB

AB = x + 6

AC = 5x - 6

BC = (5x - 6) - (x + 6)

We can kinda use distributive property, like this:

BC = 5x - 6 -x - 6

Combine like terms:

BC = 4x - 12

To verify, we can add AB and BC to see if their sum is equal to AC.

(x + 6) + (4x - 12) = 5x - 6

If we are correct, this equation will be true.

x + 6 + 4x - 12 = 5x - 6

Combine like terms:

6 + 5x - 12 = 5x - 6

5x - 6 = 5x - 6

It's correct!

BC = 4x - 12. Final Answer.

Let me know if you have any questions about this! :)

7 0

2 years ago

-3×1/4-6 please help im losing my mind

lora16 [44]

Answer:

-45/4 or -11.25

I'm pretty sure this is right hope it helps

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2 years ago

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Gordon types 1,922 words in 31 minutes. Find the unit rate

Kisachek [45]

Answer:

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Step-by-step explanation:

Hope that helps!

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1 year ago

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Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

2 years ago