What is the median of the following numbers? 1,2,10,6,4,4,6,3,1,4
2 answers:
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Answer:
a) 4 -
b) 1 -
c) 6 -
Step-by-step explanation:
It simply asks the steps to go from the original displacement formula to isolate a (the acceleration). It's just a matter of moving items around.
We start with:
We then move the vt part on the left side, then multiply each side by -1 (to get rid of the negative on the at side and to match answer choice #4):
Then we multiply each side by 2 to get rid of the 1/2, answer #1:
Finally, we divide each side by t^2 to isolate a (answer #6):
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.