<span>y = tan^−1(x2/4)</span>
tan(y) = x2/4
sec2(y) = x/2
y′ = xcos^2(y)/2
<span>cos^2(y) = <span>16x2+16</span></span>
<span>y′ = <span>8x/(<span>x2+16)
let u be x2+16
du is 2x dx
dy = 4 du / u
y = 4 ln (</span></span></span>x2 <span>+ 16)
y at x =0 = </span> 4 ln (<span>16) = 11.09</span>
Answer:
a)
b) For this case since the value 32.5 is in the confidence interval obtained then we can't conclude that the statement by Nielsen is wrong
Step-by-step explanation:
Information given
represent the sample mean
population mean (variable of interest)
s=23.1 represent the sample standard deviation
n=75 represent the sample size
Part a
The confidence interval for the mean is given by the following formula:
(1)
The degrees of freedom are given by:
Since the Confidence is 0.95 or 95%, the value of and and the critical value would be
Now we have everything in order to replace into formula (1):
Part b
For this case since the value 32.5 is in the confidence interval obtained then we can't conclude that the statement by Nielsen is wrong
Answer:
$34.77
Step-by-step explanation:
I believe that the answer is C sorry if it is wrong.