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3 years ago

The answer needs to be a mixed number a integer or a proper fraction.

Mathematics

1 answer:

kobusy [5.1K]3 years ago

7 0

Answer:

sorry sir i really dont even understand

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Round 5.456 to the nearest tenth.

docker41 [41]

Answer:

5.5

Step-by-step explanation:

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2 years ago

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12q - 6 + 7 - 3q + q

s344n2d4d5 [400]

Answer:

10q - 1

Step-by-step explanation:

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2 years ago

(1.5 x 10^2) X (2.0 x 10^3) Express your answer in scientific notation and standard form.

zavuch27 [327]

Answer:

3*10^5 or 300,000 cant remember witch one it is

Step-by-step explanation:

1.5*10^2= 150

2*10^3=2,000

150*2000=300,000

3*10^5

8 0

2 years ago

The length of an edge of a cube is 2 inches. How many cubic inches are there in the volume of the cube ?

gizmo_the_mogwai [7]

A cube has equal edges, so 2*2*2 or 2^3 (2 cubed) v=8^3

4 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