The value of z score for the history class test of Opal’s test score is negative one (-1).
<h3>What is normally distributed data?</h3>
Normally distributed data is the distribution of probability which is symmetric about the mean.
The mean of the data is the average value of the given data. The standard deviation of the data is the half of the difference of the highest value and mean of the data set.
The z score for the normal distributed data can be given as,

The mean value of the scores of history class is,

The standard deviation value of the scores of history class is,

Here, the test score of the Opal’s is 72. Thus the z score for her test score can be given as,

Hence, the value of z score for the history class test of Opal’s test score is negative one (-1).
Learn more about the normally distributed data here;
brainly.com/question/6587992
Answer:0.3, 0.386,0.683,0.836
Step-by-step explanation:
You look left to right in the numbers (0.12,0.12)
Answer:
Step-by-step explanation:
|3x+12|
one is as it is one is the negative value of it
3x + 12 - (3x + 12)
- 3x - 12
That is because both negative and positive values put in am absolute value become positive.
Answer:
13
Step-by-step explanation:
Purchase 2 upper deck tickets:

You can then buy 13 right field bleacher tickets before you reach deficit:

Because you cannot purchase a .583 of a ticket, the most you can buy is 13.
Answer:
Second choice:


Fifth choice:


Step-by-step explanation:
Let's look at choice 1.


I'm going to subtract 1 on both sides for the first equation giving me
. I will replace the
in the second equation with this substitution from equation 1.

Expand using the distributive property and the identity
:




So this not the desired result.
Let's look at choice 2.


Solve the first equation for
by dividing both sides by 2:
.
Let's plug this into equation 2:



This is the desired result.
Choice 3:


Solve the first equation for
by adding 3 on both sides:
.
Plug into second equation:

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:



Not the desired result.
Choice 4:


I'm going to solve the bottom equation for
since I don't want to deal with square roots.
Add 3 on both sides:

Divide both sides by 2:

Plug into equation 1:

This is not the desired result because the
variable will be squared now instead of the
variable.
Choice 5:


Solve the first equation for
by subtracting 1 on both sides:
.
Plug into equation 2:

Distribute and use the binomial square identity used earlier:



.
This is the desired result.