Answer:
The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 10 - 1 = 9
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of
. So we have T = 3.25
The margin of error is:
M = T*s = 3.25*25 = 81
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 12,556 - 81 = 12,475 pounds
The upper end of the interval is the sample mean added to M. So it is 12,556 + 81 = 12,637 pounds.
The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.
Answer:
Step-by-step explanation:
1/16
2^-2 * 2^-2
2^2 * 2^-6
1/4^-2
Answer:
b+6
Problem:
If the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?
Step-by-step explanation:
We are given (b+c)/2=8 and d=3b-4.
We are asked to find (c+d)/2 in terms of variable, b.
We need to first solve (b+c)/2=8 for c.
Multiply both sides by 2: b+c=16.
Subtract b on both sides: c=16-b
Now let's plug in c=16-b and d=3b-4 into (c+d)/2:
([16-b]+[3b-4])/2
Combine like terms:
(12+2b)/2
Divide top and bottom by 2:
(6+1b)/1
Multiplicative identity property applied:
(6+b)/1
Anything divided by 1 is that anything:
(6+b)
6+b
b+6
2(2x-1) + 2(3x)=4x-2+6x = 10x - 2 & not <span>=10x-1</span>