How do I fill out the table help me answer it but try to explain.

Find the z-scores that bound the middle 74% of the area under the standard normal curve.

aniked [119]

Since 74% of the middle area is bounded, this means that there is 13% on the left side, and another 13% on the right side.

P (left) = 0.13

P (right ) = 1 - 0.13 = 0.87

At this P values, the z scores are approximately:

z score (left) = -2.22

<span>z score (right) = 1.13</span>

5 0

3 years ago

What is true of a sequence of transformations that rotates an image and then translates it in order to map it onto another image

joja [24]

The true statement about the sequence of transformations is it includes exactly two rigid transformations.

<h3>How to determine the true statement?</h3>

The transformation statement is given as:

a sequence of transformations that rotates an image and then translates it in order to map it onto another image

This can be split as follows:

A sequence of transformations that rotates an image
Then translates it in order to map it onto another image

Translation and rotation are rigid transformations

This means that the size and the angle of the shape that is transformed will remain the same

Hence, the true statement about the sequence of transformations is it includes exactly two rigid transformations.

Read more about transformation at

brainly.com/question/4289712

#SPJ1

4 0

2 years ago

A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and

natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: $E=k\frac{q}{r^{2} }$ ; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: $p=\frac{q}{A}$ ; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: $S=4\pi r^{2}$ ; which gives us the surface of a sphere.