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3 years ago

I’ll mark Brainly! Which problem is being modeled on the number line?

Mathematics

2 answers:

natulia [17]3 years ago

8 0

Answer:

I think it's B.

Step-by-step explanation:

The number line shows that when

$-4-6$ is executed,

$-10$ is the answer. The answer that displays this in this order is B.

Lesechka [4]3 years ago

3 0

Answer:

The answer is B

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4 thousandths times 3 (like this plz 0.004x3)

Ksju [112]

Answer:

0.012

Step-by-step explanation:

0.004 * 3 = 0.012

8 0

3 years ago

Help!!!!!!!!!!!!!!!!!!1

Veseljchak [2.6K]

What do you need help with

8 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

Six donuts cost a total of $3.54 what is the cost for one donut? Six donuts cost a total of $3.54 what is the cost of one donut?

UkoKoshka [18]

Answer:

one doughnut costs 59 cents

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The sum of a number and 7 algebraic expression

Arisa [49]

Since it says the sum of any number not one in specific then the value will be x, plus the sum of 7,

$x + 7$

5 0

3 years ago