Woah that looks hard and if no one answers just search it up
Answer: The percent higher is 41.68%. If 49 planes were selected, 20 of them should be above 15 years.
To find the percent, we first need the z-score.
(15 - 13.5) / 7.3 = 0.21
Now, use a normal distribution table to find the percent above a score of 0.21. It will be 41.68%.
To find the number of 49 planes above this value, multiply 49 by 0.4168. You will have about 20.4 planes.
To determine the answer, we need to express the fractions into its decimal counterparts. 3/5 is equal to 0.6 therefore it is closest to 0.5 while 1/10 is equal to 0.1 therefore it is the last option. Hope this answers the question. Have a nice day.
<span>What is the next step in the proof? Choose the most logical approach. Statement: ∠1≅∠8 and ∠2≅∠7 Reason: Congruent Supplements Theorem Statement: m∠3+m∠4=180° and m∠7+m∠8=180° Reason: Linear Pair Theorem Statement: m∠3+m∠5=180° and m∠4+m∠6=180° Reason: definition of supplementary angles Statement: ∠7≅∠6 and ∠8≅∠5 Reason: Vertical Angles Theorem Done </span>
Answer:
a) H0: 
H1: 
b) 
And the critical values with 
on each tail are:
c) ![t=(10-1) [\frac{1.186}{1.34}]^2 =7.05](https://tex.z-dn.net/?f=%20t%3D%2810-1%29%20%5B%5Cfrac%7B1.186%7D%7B1.34%7D%5D%5E2%20%3D7.05)
d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34
Step-by-step explanation:
Information provided
n = 10 sample size
s= 1.186 the sample deviation
the value that we want to test
represent the p value for the test
t represent the statistic (chi square test)
significance level
Part a
On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:
H0:
H1:
The statistic is given by:
![t=(n-1) [\frac{s}{\sigma_o}]^2](https://tex.z-dn.net/?f=%20t%3D%28n-1%29%20%5B%5Cfrac%7Bs%7D%7B%5Csigma_o%7D%5D%5E2%20)
Part b
The degrees of freedom are given by:
And the critical values with on each tail are:
Part c
Replacing the info we got:
Part d
For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34
