For 4 4 to 5 16 to 20 24 to 30
Answer:
Step-by-step explanation:
A. The ratio of lynx to mountain lions to wolverines is 2:3:1.
Thus, there are 2 lynx in the ratio. If there were 6 lynx, we would have to multiply all of the numbers in the ratio by 3 (because 6/2 = 3) to keep the ratio in the same proportion.
Therefore, because there is 1 wolverine in the ratio, and 1 * 3 = 3, if there were 6 lynx, there would be 3 wolverines.
b. We can use the same ideas that we had in part a to help us in part b.
There are 3 mountain lions in the ratio, but there are 15 mountain lions in the problem. Thus, the multiplier is 5, because 15/3=5.
Therefore, because there are 2 lynx in the ratio, and 2*5 = 10, if there were 15 mountain lions, there would be 10 lynx.
c. There is one wolverine in the ratio, but there are 10 wolverines in the problem. Thus, the multiplier is 10, because 10/1 = 10
Therefore, because there are 3 mountain lions in the ratio, and 3 * 10 = 30, if there were 10 wolverines in the park, then there would be 30 mountain lions.
d. The total number of lynx, mountain lions, and wolverines is 30.
To find out how many of each animal there should be, we must make an equation using the ratio and the variable x.
2x + 3x + 1x = 30
This equation means that the total number of animals together is 30, which is true. Now let's simplify by combining like terms.
6x = 30
Finally, we can simplify by dividing both sides by the coefficient of x, or 6.
x = 5
Thus, going back to our original equation, we know that the amount of lynx is 2x, mountain lions is 3x, and wolverines is 1x.
Lynx = 2x = 2(5) = 10 lynx
Mountain Lion = 3x = 3(5) = 15 mountain lions
Wolverines = 1x = 1(5) = 5 wolverines
Hope this helps! :)
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.