A pair of dice is rolled. List the elements of sample space S. Let A denote the event “the sum is less than 5” and B the event “
1 answer:
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Answer:
Part A: The null hypothesis failed to be rejected.
Part B: The null hypothesis failed to be rejected.
Step-by-step explanation:
We have an hypothesis test with null and alternative hypothesis H0: p = 0.5 versus Ha: p > 0.5, which has the test statistic z=1.15.
Part A: If the significance level is 0.05, the conclusion depends on the P-value.
If the P-value is below 5%, the null hypothesis is rejected.
The P-value for this right-tailed tes and z=1.15 is:
The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.
Part B: In this case, the significance level is 0.01 and, as the alternative hypothesis is defined with an unequal sign, the test is two-tailed.
This changes the way we calculate the P-value, as we need to compute the two tails.
The P-value is:
The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.
The equation -21t - 4 = 10 - 2t has only one solution
<u>Solution:</u>
Given, equation is – 21t – 4 = 10 – 2t
We have to find how many solutions does the above given equation have
Now, let us solve the given equation to find the number of solutions it has.
Then, - 21t – 4 = 10 – 2t
Taking like terms to one side of the equation we get,
- 21t + 2t = 10 + 4
here we can see only one value of t.
Hence, the given equation has only one solution.
The answers are:
1)
2)
3)
The explanation for this answer is shown below.
1. To solve this problem you must apply the following proccedure:
2. You have the function
3. You must switch and
and then solve for
:
4. Finally, you obtain: