4/5 is already simplified so its going to be 4/5 or .45
Answer:
r = -5. Find the pattern and write the recursive formula: a_n + 1 = - 5 a_nGiven the recursive formula: r =
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Answer with explanation</u>
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Let
be the population mean.
As per given , we have

Since the alternative hypothesis is right-tailed , so the test is a right-tailed test.
Also, population standard deviation is given
, so we perform one-tailed z-test.
Test statistic : 
, where
= Population mean
= Population standard deviation
n= sample size
= Sample mean
For n= 18 ,
,
,
, we have

P-value (for right tailed test): P(z>2.12) = 1-P(z≤ 2.12) [∵ P(Z>z)=1-P(Z≤z)]\
=1- 0.0340=0.9660
Decision : Since P-value(0.9660) > Significance level (0.01), it means we are failed to reject the null hypothesis.
[We reject null hypothesis if p-value is larger than the significance level . ]
Conclusion : We do not have sufficient evidence to show that the goal is not being met at α = .01 .
First you find 4/5 of 8 and then 4/5of 7 and multiply them
The remaining of them is to get the math app can’t do it for me