Answer:
Step-by-step explanation:
Answer:
Step-by-step explanation:
Tthe population is normally distributed and the sample size is .
Since the population standard deviation is unknown and the sample standard deviation , must replace it, the t distribution must be used for the confidence interval.
Hence with degrees of freedom of 21, .(Read from the t distribution table)
The 98% confidence interval can be constructed using the formula:
.
From the question the sample mean is dollars and the sample standard deviation is dollars.
We substitute the values into the formula to get
Therefore, we can be 98% confident that the population mean is between is between 19.22 and 44.78 dollars.
Answer: x= -10
Step-by-step explanation:
-3/4*(x+2) = 6 /multiply both sides by (-4/3)
-4/3*(-3/4)*(x+2) = 6*(-4/3) /-4/3*(-3/4) = 1
x+2 = -24/3 /-24/3 = -8
x+2 =-8 /subtract 2 from both sides
x = -8-2
x = -10
Answer:
(a)
(b)
Step-by-step explanation:
Let's call the length of the box L, the width W and the height H. Then, we can write the following equations:
"A rectangular box has a base that is 4 times as long as it is wide"
"The sum of the height and the girth of the box is 200 feet"
The volume of the box is given by:
Using the L and H values from the equations above, we have:
The domain of V(W) is all positive values of W that gives a positive value for the volume (because a negative value for the volume or for the width doesn't make sense).
So to find where V(W) > 0, let's find first when V(W) = 0:
The volume is zero when W = 0 or W = 100.
For positive values of W ≤ 100, the term W^2 is positive, but the term (W - 100) is negative, then we would have a negative volume.
For positive values of W > 100, both terms W^2 and (W - 100) would be positive, giving a positive volume.
So the domain of V(W) is W > 100.
By the Stolz-Cesaro theorem, this limit exists if
also exists, and the limits would be equal. The theorem requires that
be strictly monotone and divergent, which is the case since
.
You have
so we're left with computing
This can be done with the help of Stirling's approximation, which says that for large
,
. By this reasoning our limit is
Let's examine this limit in parts. First,
As
, this term approaches 1.
Next,
The term on the right approaches
, cancelling the
. So we're left with
Expand the numerator and denominator, and just examine the first few leading terms and their coefficients.
Divide through the numerator and denominator by
:
So you can see that, by comparison, we have
so this is the value of the limit.