The cost of having a package delivered has a base fee of $10.80. Every pound over 5 lbs cost an additional $0.41 per pound. Writ
2 answers:
Answer:
The required equation is .
Step-by-step explanation:
It is given that the cost of having a package delivered has a base fee of $10.80. Every pound over 5 lbs cost an additional $0.41 per pound.
Fixed cost is $10.80.
Variable cost is $0.41w, where w is the total weight in pounds of a package weighing over 5 lbs.
Total cost is the addition of fixed cost and variable cost.
Where, C is expresses the cost C in terms of w, the total weight in pounds of a package weighing over 5 lbs.
Therefore the required equation is .
You might be interested in
Answer:
<h2> aₙ = 7n - 10</h2>Step-by-step explanation:
The nth term:
So:
a₄ = a₁ + 3d
18 = a₁ + 3d ⇒ 3d = 18 - a₁
a₇ = a₁ + 6d
39 = a₁ + 6d
39 = a₁ + 2(18 - a₁)
39 = a₁ + 36 - 2a₁
39 = 36 - a₁
a₁ = -3
3d = 18 - (-3)
3d = 21
d = 7
So the nth term:
Answer:
a)
b)
c) On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of the margin of error change for part a and b.
Step-by-step explanation:
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
The margin of error for the proportion interval is given by this formula:
(a)
If solve n from equation (a) we got:
(b)
Part a
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by and
. And the critical value would be given by:
If we replace the values into equation (a) for 1983 we got:
Part b
Since is the same confidence level the z value it's the same.
If we replace the values into equation (a) for 2008 we got:
Is the margin of error the same in parts (a) and (b)? Why or why not?
On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of the margin of error change for part a and b.