Answer:
Correct option: B
Step-by-step explanation:
The professor can perform a One-mean <em>t</em>-test to determine whether the average score of the students in his class is more than the average score of all the students attending university.
A <em>t</em>-test will be used instead of the <em>z</em>-test because the population standard deviation is not provided instead it is estimated by the sample standard deviation.
The hypothesis for this test can be defined as follows:
<em>H₀</em>: The average score of the students in his class is not more then the entire university, i.e. <em>μ ≤ 35</em>.
<em>Hₐ</em>: The average score of the students in his class is more then the entire university, i.e. <em>μ > 35</em>.
Given:

The test statistic is:

Thus, the correct option is (B).
Answer:
Volume = 144 ft
Step-by-step explanation:
1. Let's write some things down, so we don't forget them!
Length = 12 ft
Height = 2 ft
Width = 6 ft
2. We can probably assume the sandbox is a rectangle. The formula for the volume of a rectangle is Length x Height x Width. So, 12 x 2 x 6 = 144, therefore that's the volume!
Using the normal distribution, it is found that:
a) The pilot is at the 72th percentile.
b) 19.13% of pilots are unable to fly.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
.
Item a:
The percentile is the <u>p-value of Z when X = 74.2</u>, hence:


Z = 0.59
Z = 0.59 has a p-value of 0.7224.
72th percentile.
Item b:
The proportion that is able to fly is the <u>p-value of Z when X = 78 subtracted by the p-value of Z when X = 70</u>, hence:
X = 78:


Z = 2
Z = 2 has a p-value of 0.9772.
X = 70:


Z = -0.96
Z = -0.96 has a p-value of 0.1685.
0.9772 - 0.1685 = 0.8087 = 80.87%.
Hence the percentage that is unable to fly is:
100 - 80.87 = 19.13%.
More can be learned about the normal distribution at brainly.com/question/4079902
#SPJ1
Answer:
-129
Step-by-step explanation:
2x+5=0
x=-5/2 [By the remainder theorem]
f(-5/2)=8(-5/2)^3+4(-5/2)^2+13(-5/2)+3
=-125+25-65/2+3
=-129.
(-8)x=(-6)x+10
(-2)x=10
X=10/-2
X=-5