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djverab [1.8K]
3 years ago
13

An object starts at position 12 on a horizontal line with a reference point of 0. What is the position of the object if it

Mathematics
2 answers:
Dahasolnce [82]3 years ago
8 0

Answer:

-2

Step-by-step explanation:

We know that the object is at 12 on a horizontal line.

The horizontal line of a graph is the x axis, so lets think of 12 as 12x

On a graph, left is negative, and right is postive.

In this case we need to move 14x to the right, or -14x.

We start at 12x, so its going to be:

12x-14x=-2x

So -2 is the postition of the object on the x axis.

This is answer B.

Hope this helps!

poizon [28]3 years ago
3 0

Answer:

The answer would be B

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Answer:

E(X) = 7*0.18 +9*0.08 +11*0.09 +15*0.08 +17*0.41 =13.22

And we can find the second moment with this formula:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 7^2*0.18 +9^2*0.08 +11^2*0.09 +15^2*0.08 +17^2*0.41 =189.72

And we can find the variance like this:

Var(X) = E(X^2) -[E(X)]^2= 189.72- (13.22)^2 =14.9516

And the deviation would be:

Sd(X)= \sqrt{14.9516}= 3.867

Step-by-step explanation:

For this case we have the following dataset given:

Payment     $7     $9     $11    $13    $15  $17

Probability 0.18  0.08  0.09  0.16  0.08   0.41

For this case we can calculate the mean with this formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(X) = 7*0.18 +9*0.08 +11*0.09 +15*0.08 +17*0.41 =13.22

And we can find the second moment with this formula:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 7^2*0.18 +9^2*0.08 +11^2*0.09 +15^2*0.08 +17^2*0.41 =189.72

And we can find the variance like this:

Var(X) = E(X^2) -[E(X)]^2= 189.72- (13.22)^2 =14.9516

And the deviation would be:

Sd(X)= \sqrt{14.9516}= 3.867

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