Consider rectangle PQRS (not shown) with PQ=12 and PS=16. If A and B are the midpoints of sides PQ?????? and QR?????? respective
ly, find AB.
1 answer:
Answer: The answer is AB =10
Step-by-step explanation:
In the given figure there is a rectangle PQRS
Given :PQ=12
PS=16
Since it is a rectangle
∵ PQ=RS
and PS=QR
i.e. QR=16
A is mid point of PQ
⇒QA =
=6
B is mid point of QR
⇒ QB=
=8
So In Δ AQB
Applying pythagoras theorem
AQ² +BQ² =AB²
∵ AB²=6²+8²
=36 +64
=100
i.e. AB=√100
=10
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Answer:
Null hypothesis:
Alternative hypothesis:
b) 2.70
b. 1.85
b. 0.03
a. We can reject H(0) in favor of H(a)
Step-by-step explanation:
Data given and notation
represent the mean for the sample 1
represent the mean for the sample 2
represent the population standard deviation for the sample 1
represent the population standard deviation for the sample B2
sample size selected 1
sample size selected 2
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the expenditure of households in City 1 is more than that of households in City 2, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".
Standard error
The standard error on this case is given by:
Replacing the values that we have we got:
b. 2.70
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
Since is a one side right tailed test the p value would be:
b. 0.03
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis.
a. We can reject H(0) in favor of H(a)