What is the number has 8 zeros in it

Nichelle earns $7 per hour and gets a 10% commission on the sale price of each item she sells. She wants to work only 10 hours e

kirza4 [7]

Answer:

no sé amigo apenas voy el el kinder

8 0

2 years ago

Drag each equation to show if it could be a correct first step to solving the equation 2(x+7)=36.

DanielleElmas [232]

The second is Yes
Sixth is Yes
First is No
Fourth is No
Fifth is No
Third is Not Enough Info
And last x=11

7 0

2 years ago

Read 2 more answers

Use the graph to answer the question.

Lelu [443]

Answer:

Coordinate Q is (0.8, 0.7)

Step-by-step explanation:

We are told that the coordinates of point Pare (0.6,0.1).

This means that along the x-axis, x = 0.6 and along the y-axis, y = 0.1.

Now, by inspection of the graph, we can see that when we count boxes from the origin to the point P, we have 6 boxes. Thus, each box corresponds to 0.1. So, for point Q, from the origin to that point, on the x-axis, we have 8 boxes. Since one box = 0.1, then the x - value of Coordinate Q is 0.8.

On the y - axis, we see that we have one box from the origin up for the corresponding y-value of coordinate P.

This means that one box is 0.1.

For coordinate Q, we will count 7 boxes. Thus, y-value of coordinate Q is 0.7.

Thus,coordinate Q is (0.8, 0.7)

7 0

2 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

Jace invested $380 in an account paying an interest rate of 6.2% compounded continuously. Assuming no deposits or withdrawals ar

SOVA2 [1]

Answer: 960

Step-by-step explanation:

5 0

2 years ago