A girls' track team must run 3 miles on the first day of practice and 6 miles every day after that. The boys' team must run 5 mi
2 answers:
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Answer:
We need a sample size of at least 719
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?
This is at least n, in which n is found when . So
Rouding up
We need a sample size of at least 719
<h2>Answer:</h2>
In the 5th year
<h2>Step-by-step explanation:</h2>For the first year, the salary is 1.2million = 1,200,000
For the second year, the salary is 1.2million + 75000 = 1,200,000 + 75,000 = 1,275,000
.
.
.
For the last year, the salary is 1.5million = 1,500,000
This gives the following sequence...
1,200,000 1,275,000 . . . 1,500,000
This follows an arithmetic progression with an increment of 75,000.
<em>Remember that,</em>
The last term, L, of an arithmetic progression is given by;
L = a + (n - 1)d ---------------(i)
<em>Where;</em>
a = first term of the sequence
n = number of terms in the sequence (which is the number of years)
d = the common difference or increment of the sequence
<em>From the given sequence,</em>
a = 1,200,000 [which is the first salary]
d = 75,000 [which is the increment in salary]
L = 1,500,000 [which is the maximum salary]
<em>Substitute these values into equation (i) as follows;</em>
1,500,000 = 1,200,00 + (n - 1) 75,000
1,500,000 - 1,200,000 = 75,000(n-1)
300,000 = 75,000(n - 1)
4 = n - 1
n = 5
Therefore, in the 5th year the maximum salary will be reached.