Answer:
904.32inc^3
Step-by-step explanation:
<em>S</em><em>E</em><em>E</em><em> </em><em>I</em><em>M</em><em>A</em><em>G</em><em>E</em><em> </em><em>F</em><em>O</em><em>R</em><em> </em><em>A</em><em>N</em><em>S</em><em>W</em><em>E</em><em>R</em><em>.</em><em>.</em><em>.</em>
Answer:
<2 = 129°
Step-by-step explanation:
if I || m then 5x+12 =m<2 and m<2 + 2x + 14 = 180°
2x + 14 + 5x + 12 = 7x + 26 =180°
7x = 154 ➡x = 21°
m <2 = 5×21 + 14
Distribute the five to [3q-10]
Combine like terms [15q] and [-5q]
Add 50 on each side
Divide by 10 on each side
Simplify
Answer:
16.7%
Step-by-step explanation:
Please check below attachment for sample space table
Firstly, we have the total number of possible results as 12
From the 12, the probability of flipping a head is 6/12
Rolling either a 3 or a 5 is the sum of places we have both divided by the possible total of 12
we have this as ; 2/12 + 2/12 = 4/12
The term ‘AND’ means we have to multiply the two probabilities as follows;
4/12 * 1/2 = 1/6 = 0.1667
To the nearest tenth, we have this as 16.67% which is 16.7%
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