Answer:
D. 67
Step-by-step explanation:
To help me see the region we needed to look at, I wrote the inequalities out.
Let x be number of pre-sale tickets and y be the number of at-the-door tickets as your graph suggests.
So one of the inequalities about number of where the other one is about cost.
You are given x+y is no more than 400 or x+y<=400 (the top line graphed in your picture is x+y=400).
You are given 10x+25y is at least 5000 or 10x+25y>=5000 (the bottom line graphed in your picture).
I solved both of these for y.
x+y<=400
Subtract x on both sides giving y<=-x+400 (shaded below line because of the y< part).
10x+25y>=5000
Subtract 10x on both sides:
25y>=-10x+5000
Divide both sides by 25:
y>=-2/5 x+200 (shaded above the line because of y> part).
The region we should then be looking at is:
Let's look at the points (0,400), (0,200), and finally (333,67).
Cost=10x+25y
Let's plug in
Cost=10(0)+25(400)=10000
Cost=10(0)+25(200)=5000 (can we go lower than 200)
Cost=10(333)+25(67)=5005 (our y is lower 200 so far this is the winner)
Cost 10(334)+25(66)=4990 (didn't meet the 5000 dollar requirement)
67 D.
(Also if you look at the graph 66 would not be included in the shaded region; it would be too low)
Answer:
We failed to reject H₀
Z > -1.645
-1.84 > -1.645
We failed to reject H₀
p > α
0.03 > 0.01
We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.
Step-by-step explanation:
Set up hypotheses:
Null hypotheses = H₀: p = 0.40
Alternate hypotheses = H₁: p < 0.40
Determine the level of significance and Z-score:
Given level of significance = 1% = 0.01
Since it is a lower tailed test,
Z-score = -2.33 (lower tailed)
Determine type of test:
Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.
Select the test statistic:
Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.
Set up decision rule:
Since it is a lower tailed test, using a Z statistic at a significance level of 1%
We Reject H₀ if Z < -1.645
We Reject H₀ if p ≤ α
Compute the test statistic:
From the z-table, the p-value corresponding to the test statistic -1.84 is
p = 0.03288
Conclusion:
We failed to reject H₀
Z > -1.645
-1.84 > -1.645
We failed to reject H₀
p > α
0.03 > 0.01
We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.
Answer:
Step-by-step explanation:
A picture of the question is shown in the figure attached. There we can see that the step corresponding to Properties of exponents is the option:
which is equivalent to
