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3 years ago

What is the value of x ? please explain ! ill mark brainlest !!

Mathematics

1 answer:

Sedaia [141]3 years ago

3 0

Answer

ehfiefewfiehfwenfkwe good luck

Step-by-step explanation:

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12 3/8 divided by 3/4

oksano4ka [1.4K]

Answer:

32/2 or 16.5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

47 was divided by the power of ten to get 0.047 what power of ten was divided

Dmitrij [34]

Answer:

1,000

Step-by-step explanation:

47 to 0.047 moves 3 places

47, 4.7, 0.47, 0.047

this means that the power of then used should have 3 0's

making the power of ten 1,000

3 0

3 years ago

You buy a new car for $31,300. The average car depreciates in value by 14% per year. How much will it be worth after each of the

Andrei [34K]

A = P(1 + rt)

A=31,300(1+.14*1)

A=35682

5 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

What is the interquartile range of this data? A.6 B.7 C.8 D.9

Butoxors [25]

Hello!

To find the interquartile range, subtract the value of the upper quartile from the value of the lower quartile.

33 - 25 = 8

The interquartile range is 8. Hope I helped! :3

6 0

3 years ago