Given:
Accuracy = 5
99% confidence interval
s = 17, sample standard deviation.
Because the population standard deviation is unknown, we should use the Student's t distribution.
The accuracy at the 99% confidence level for estimating the true mean is

where
n = the sample size.
t* is provided by the t-table.
That is,
(17t*)/√n = 5
√n = (17t*)/5 = 3.4t*
n = 11.56(t*)²
A table of t* values versus df (degrees of freedom) is as follows.
Note that df = n-1.
n df t*
------ -------- -------
1001 1000 2.581
101 100 2.626
81 80 2.639
61 60 2.660
We should evaluate iteratively until the guessed value, n, agrees with the computed value, N.
Try n = 1001 => df = 1000.
t* = 2.581
N = 11.56*(2.581²) = 77
No agreement.
Try n = 81 => df = 80
t* = 2.639
N = 11.56*(2.639²) = 80.5
Good agreement
We conclude that n = 81.
Answer: The sample size is 81.
Answer:
The x coordinate changes
Step-by-step explanation:
When you are reflecting a coordinate on the y-axis, the x-axis to becomes negative, though the y-axis stays the same.
Answer:
I cant see the picture clearly, so I cant help you :( Sorry
Step-by-step explanation:
Step-by-step explanation:
1/2, 2, 9/2, ....
equivalent with :
½, 4/2, 9/2, ...
the rule is : Un = n²/2
so, U10= 10²/2 = 100/2= 50
<span>The answer is: the least amount is 105.35 and the greatest amount is 105.44. If the number after the one you want to round is 5 or bigger, you need to round up the number. For instance, 105.35 is rounded to 105.4 which is equal to 105.40. If the number after the one you want to round is smaller than 5, you need to round down the number. For instance, 105.44 is rounded to 105.4 which is equal to 105.40.</span>