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3 years ago

5

Which model shows the problem 1 ÷ 1 6 ? A pie chart with 6 shaded parts. A solid circle. 3 ovals each containing 2 circles. 2 ov

als each contained 3 circles.

2 answers:

nadya68 [22]3 years ago

4 0

Answer:

i think its the one about pie charts, but im not sure.

Step-by-step explanation:

3241004551 [841]3 years ago

4 0

The one about pie charts

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The median of the numbers would be 7.5

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3 years ago

Which of the following is the equation for the graph shown?

serious [3.7K]

The Answer is A

Explanation:

The standard form equation for an ellipse is:

(x - h)² / a² + (y - k)² / b² = 1

So looking at the picture above you can see that the center of the ellipse is at the origin which is (0;0) making both h and k zero.

It is then observed that the x-intercepts are at 12 meaning a=12 and y- intercepts are at approximately square root of 95 which is 9.746... which are 'k'

So you can conclude by saying that:

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3 years ago

Tyler’s brother earns a commission. He makes 2.5% of the amount he sells. Last week, he sold $900 worth of shoes. How much was h

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Step-by-step explanation:

2.5% of 900

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3 years ago

Suppose that a college determines the following distribution for X = number of courses taken by a full-time student this semeste

lidiya [134]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$ In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$ And the variance is given by:

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And the deviation would be:

$Sd(X) =\sqrt{0.8924} =0.9447$

Step-by-step explanation:

Previous concepts

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).

Solution to the problem

For this case we have the following distribution given:

X 3 4 5 6

P(X) 0.07 0.4 0.25 0.28

We can calculate the mean with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$

In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$

And the variance is given by:

$Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924$

And the deviation would be:

$Sd(X) =\sqrt{0.8924} =0.9447$

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3 years ago

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sergij07 [2.7K]

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3 years ago