Solve this 5a-d please
1 answer:
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Answer:
The 99% of a confidence interval for the average maximum HP for the experimental engine.
(536.46, 603.54)
Step-by-step explanation:
<u><em>Step:-1</em></u>
Given that the mean of the Population = 540HP
Given that the size of the sample 'n' = 9
Given that the mean of the sample = 570HP
Given that the sample standard deviation = 30HP
<u><em>Step(ii):-</em></u>
<u><em>Degrees of freedom = n-1 =9-1 =8</em></u>
<u><em>t₀.₀₀₅ = 3.3554</em></u>
The 99% of a confidence interval for the average maximum HP for the experimental engine.
(570 - 33.54 , 570+33.54)
(536.46 , 603.54)
<u><em>Final answer</em></u> :-
<em>The 99% of a confidence interval for the average maximum HP for the experimental engine.</em>
<em>(536.46, 603.54)</em>
<em></em>
Answer:
There are 80 marks in total.
Step-by-step explanation:
Let the number of total marks be .
The percentage score of the student can be written as the ratio
.
However,
.
Equating the two:
.
Cross-multiply (that is: multiple both sides by , the product of the two denominators) to get
.
.
In other words, there are 80 marks in total.