In the diagram, JOK and MON are straight lines. Angle NOP and Angle NOQ are complimentary angles. Calculate the value of x - y.
1 answer:
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Answer:
The degreess of freedom are given by:
Now we can calculate the p value with the following probability:
For this case since the p value is higher than the significance level we can FAIL to reject the null hypothesis and we can conclude that the true means are NOT significantly different at 10% of significance.
Step-by-step explanation:
Information provided
represent the mean for sample 1
represent the mean for sample 2
represent the sample standard deviation for 1
represent the sample standard deviation for 2
sample size for the group 2
sample size for the group 2
Significance level provided
t would represent the statistic
Hypothesis to test
We want to verify if the true means for this case are significantly different, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
The statistic is given by:
(1)
And the degrees of freedom are given by
Replacing the info given we got:
The degreess of freedom are given by:
Now we can calculate the p value with the following probability:
For this case since the p value is higher than the significance level we can FAIL to reject the null hypothesis and we can conclude that the true means are NOT significantly different at 10% of significance.
Proof of the Law of Sines
The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)
a
sin A
=
b
sin B
=
c
sin C
For more see Law of Sines.
Acute triangles
Draw the altitude h from the vertex A of the triangle
From the definition of the sine function
sin B =
h
c
a n d sin C =
h
b
or
h = c sin B a n d h = b sin C
Since they are both equal to h
c sin B = b sin C
Dividing through by sinB and then sinC
c
sin C
=
b
sin B
Repeat the above, this time with the altitude drawn from point B
Using a similar method it can be shown that in this case
c
sin C
=
a
sin A
Combining (4) and (5) :
a
sin A
=
b
sin B
=
c
sin C
- Q.E.D
Obtuse Triangles
The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.
First the interior altitude. This is the same as the proof for acute triangles above.
Draw the altitude h from the vertex A of the triangle
sin B =
h
c
a n d sin C =
h
b
or
h = c sin B a n d h = b sin C
Since they are both equal to h
c sin B = b sin C
Dividing through by sinB and then sinC
c
sin C
=
b
sin B
Draw the second altitude h from B. This requires extending the side b:
The angles BAC and BAK are supplementary, so the sine of both are the same.
(see Supplementary angles trig identities)
Angle A is BAC, so
sin A =
h
c
or
h = c sin A
In the larger triangle CBK
sin C =
h
a
or
h = a sin C
From (6) and (7) since they are both equal to h
c sin A = a sin C
Dividing through by sinA then sinC:
a
sin A
=
c
sin C
Combining (4) and (9):
a
sin A
=
b
sin B
=
c
sin C
