If x=-2 then write the ordered pair that is a solution of 5x-3y=-25

Suppose Q and R are independent events. Find P(Q and R) if P(Q) = 0.63 and P(R) = 0.48.

Oksanka [162]

Hello,

Q and R are independent if p(Q and R)=p(Q)*p(R)
Here p(Q and R)=0.63-0.48=0.3024
ANSWER B

5 0

3 years ago

Read 2 more answers

I need some help please out

IceJOKER [234]

Question:

Solution:

Let the following equation:

$\sqrt[]{12-x}=\text{ x}$

this is equivalent to:

$(\sqrt[]{12-x})^2=x^2$

this is equivalent to:

$12-x=x^2$

this is equivalent to:

$x^2+x-12=\text{ 0}$

thus, we can conclude that

x= 3.

7 0

1 year ago

Find the measure of one interior angle for the following regular polygon

shusha [124]

Answer:

Step-by-step explanation:

I am assuming that by "interior" angle you do not mean the central angle.

This is a 10-sided polygon, a decagon. That means that there are 10 triangles that can extend from the center, with their sides being equal to the radii of the decagon. If we extract one of these triangles we can find what the interior angle is. The vertex angle measures 360/10 which is 36.

Split this triangle in half from the vertex to the base, creating a right triangle. The vertex angle is also split in half, making this angle (the vertex angle is the one at the top of the triangle) 18 degrees. We already know that one angle inside this right triangle is 90 (definition of a right triangle) and to find the other one, we apply the Triangle Angle-Sum Theorem:

180 - 18 - 90 = 72 degrees. That is the measure of the base angle that is NOT the right angle, obviously.

4 0

3 years ago

Radius how do I get a radius of 4cm

patriot [66]

Radius is half of the diameter so the radius of 4 is 2

4 0

3 years ago

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Explain the circumstances for which the interquartile range is the preferred measure of dispersion. What is an advantage that th

KatRina [158]

Answer:

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

Step-by-step explanation:

Previous concepts

The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.

$IQR= Q_3 -Q_1$

The standard deviation is a measure of dispersion obatined from the sample variance and is given by:

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Solution to the problem

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

8 0

3 years ago