What percent is 272 out of 355?
1 answer:
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Answer: C) angle 2 and angle 7
Imagine the parallel lines form the boundaries of river. Angles 3, 4, 5 and 6 are considered interior angles as they are in the river. Angles 1, 2, 7, 8, 9 and 10 are exterior because they're outside the river.
Angles 2 and 7 are on opposite sides of the transversal cut that I've highlighted in red (see diagram below). So that's why they're considered alternate. Combine "exterior" and "alternate" to get "alternate exterior"
Answer:
(a) We reject our null hypothesis.
(b) We fail to reject our null hypothesis.
(c) We fail to reject our null hypothesis.
Step-by-step explanation:
We are given that a certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr.
A random sample of 18 pens is selected.
<u><em>Let </em></u><u><em> = true average writing lifetime under controlled conditions</em></u>
So, Null Hypothesis, :
10 hr {means that the true average writing lifetime under controlled conditions is at least 10 hr}
Alternate Hypothesis, :
< 10 hr {means that the true average writing lifetime under controlled conditions is less than 10 hr}
<u>The test statistics that is used here is one-sample t test statistics;</u>
T.S. = ~
where, = sample mean
s = sample standard deviation
n = sample size of pens = 18
n - 1 = degree of freedom = 18 -1 = 17
<u>Now, the decision rule based on the critical value of t is given by;</u>
- If the value of test statistics is more than the critical value of t at 17 degree of freedom for left-tailed test, then <u>we will not reject our null hypothesis</u> as it will not fall in the rejection region.
- If the value of test statistics is less than the critical value of t at 17 degree of freedom for left-tailed test, then <u>we will reject our null hypothesis</u> as it will fall in the rejection region.
(a) Here, test statistics, t = -2.4 and level of significance is 0.05.
<em>Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.</em>
Here, clearly the value of test statistics is less than the critical value of t as -2.4 < -1.74, so we reject our null hypothesis.
(b) Here, test statistics, t = -1.83 and level of significance is 0.01.
<em>Now, at 0.051 significance level, the t table gives critical value of -2.567 at 17 degree of freedom.</em>
Here, clearly the value of test statistics is more than the critical value of t as -2.567 < -1.83, so we fail to reject our null hypothesis.
(c) Here, test statistics, t = 0.57 and level of significance is not given so we assume it to be 0.05.
<em>Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.</em>
Here, clearly the value of test statistics is more than the critical value of t as -1.74 < 0.57, so we fail to reject our null hypothesis.