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3 years ago

PLEASE PLEASE PLEASE HELP MEE!!!!!!!!!!!!!!!!

Mathematics

1 answer:

ivanzaharov [21]3 years ago

5 0

Answer:

It is C

Step-by-step explanation:

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What is the answer to 2(9+7)

FrozenT [24]

Answer: 32

Step-by-step explanation:

1.) Write the initial equation: 2(9+7)

2.) Add the values in the parentheses: 2(16)

3.) Multiply the value in the parentheses by 2: 2(16) = 32

5 0

3 years ago

Read 2 more answers

14=h/5<br><br> Help me please

ValentinkaMS [17]

Answer:

14=h/5

h=14*5

h=70

<u>Check</u>:

70/5=14

Hope this helps

6 0

3 years ago

Explain how knowing 50×4 = 200 helps you find 500 times 400.

skad [1K]

it equals to 200,000

4 0

3 years ago

Percentage change 42 increased to 145

dedylja [7]

The answer should be 103? because when you add 103 to 42 u should get 103%.

3 0

3 years ago

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

3 years ago