Answer:
Option B is correct =
Step-by-step explanation:
<u>The complete question is:</u> Which of the following options have the same value as 30% of 81?
Group of choices is:
(A)
(B)
(C)
(D)
(E)
Now, the expression given to us is 30% of 81.
Simplifying the above expression we get;
30% of 81 =
= =
Now, we will solve each of the given options and then see which option matches with our calculation.
Option (A) is given;
=
This doesn't match with our answer, so this option is not correct.
Option (B) is given;
<u><em>This matches with our answer, so this option is correct.</em></u>
Option (C) is given;
This doesn't match with our answer, so this option is not correct.
Option (D) is given;
=
This doesn't match with our answer, so this option is not correct.
Option (E) is given;
This doesn't match with our answer, so this option is not correct.
Answer:
106.33
Step-by-step explanation:
after decimal you have .3 tenths 2 hundredths 7 thousandths
to round to the nearest hundredths look at thousandths 7 becasuse is greater then 5 then you round up
106.327 is rounded to 106.33
Answer: Each payment is $171
Step-by-step explanation:
2052 / 12 = 171
Each payment is $171
Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.
<h3>Normal Probability Distribution</h3>
In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation .
In this problem:
- The mean is of 660, hence .
- The standard deviation is of 90, hence .
- A sample of 100 is taken, hence .
The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:
By the Central Limit Theorem
has a p-value of 0.8665.
1 - 0.8665 = 0.1335.
0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.
To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213