Which theorem or postulate can be used to prove ΔRPQ ≅ ΔRST? Given R is the midpoint of QT and QT bisects PS at R.
2 answers:
Answer:
<h2>C. SAS</h2>Step-by-step explanation:
The graph of this problem is attached. There you can observe that we demonstrate the congruence using SAS postulate, which is Side-Angle-Side.
This is the postulate we should use because the mid point divides in equal parts by definition, that means, sides RP and RS are congruent, and sides QR and TR are congruent. In addition, angles QRP and TRS are congruent, because they are vertical angle.
So, we have Side-Angle-Side as congruent parts. Therefore, the postulate we have to use is SAS, option C.
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Answer:
So on this case the 90% confidence interval would be given by
For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
represent the sample mean 1
represent the sample mean 2
n1=15 represent the sample 1 size
n2=15 represent the sample 2 size
population sample deviation for sample 1
population sample deviation for sample 2
parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%
We want to test:
H0:
H1:
And we can do this using the confidence interval for the difference of means.
Solution to the problem
The confidence interval for the difference of means is given by the following formula:
(1)
The point of estimate for is just given by:
The degrees of freedom are given by:
Since the Confidence is 0.90 or 90%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that
Now we have everything in order to replace into formula (1):
So on this case the 90% confidence interval would be given by
For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.