Select the values that make the inequality 4q≥−68 true. Then write an equivalent inequality, in terms of q. (Numbers written in
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Part a .
A arithmetic sequence with a third term of 8 and a common difference of 5 .
To find the first five therms, since the common difference is 5, so we add 5 to get the fourth term and add 5 to fourth term to get the fifth term .
And for first two terms, we will subtract 5 from 8 to get the second term and subtract 5 from the second term to get the first term. And we will get
Part b:A geometric sequence with a fifth term of 1/3 and constant ratio of 1/3.
TO find the first five terms, since the constant ratio of 1/3, so we multiply 1/3 to third term to get fourth term, and multiply fourth term by 1/3 to get fifth term .
And to get the first two terms, we will divide third term by 1/3, to get the second term and divide the second term by 1/3 to get the first term, that is
Regarding the hypothesis tested in this problem, it is found that:
a) The test statistic is: t = 0.49.
b) The p-value is: 0.6297..
c) The null hypothesis is not rejected, as the p-value is greater than the significance level.
d) It appears that the students are legitimately good at estimating one minute, as the sample does not give enough evidence to reject the null hypothesis.
<h3>What is the test statistic?</h3>The test statistic of the t-distribution is given by the equation presented as follows:
In which the variables of the equation are given as follows:
is the sample mean.
is the value tested at the hypotheses.
- s is the standard deviation of the sample.
- n is the sample size.
The value tested at the hypothesis is:
From the sample, using a calculator, the other parameters are given as follows:
Hence the test statistic is obtained as follows:
t = 0.49.
<h3>What are the p-value and the conclusion?</h3>The p-value is obtained using a t-distribution calculator, with a two-tailed test, as we are testing if the mean is different a value, in this case 60, with:
- 15 - 1 = 14 df, as the number of degrees of freedom is one less than the sample size.
- t = 0.49, as obtained above.
Hence the p-value is of 0.6297.
Since the p-value is greater than the significance level of 0.01, the null hypothesis is not rejected, attesting to the ability of the students to estimate one minute.
More can be learned about the test of an hypothesis at brainly.com/question/13873630
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