What is the awnser to this question
1 answer:
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(a) By DeMoivre's theorem, we have
On the LHS, expanding yields
Matching up real and imaginary parts, we have for (i) and (ii),
(b) By the definition of the tangent function,
(c) Setting
, we have 
and 
. So
At the given value of
, the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.
Remember, this is saying that
If we replace
with a variable 
, then the above means is a root to the quadratic equation,
Also, if
, then 
and 
. So by a similar argument as above, we deduce that 
is also a root to the quadratic equation above.
(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write
Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy
But
for all 
, as is the case for 
and 
, so we choose the positive root.
On the LHS, expanding yields
Matching up real and imaginary parts, we have for (i) and (ii),
(b) By the definition of the tangent function,
(c) Setting
At the given value of
Remember, this is saying that
If we replace
Also, if
(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write
Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy
But
Answer:
Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795
Step-by-step explanation:
Given
See attachment for class
Solving (a): Fill the midpoint of each class.
Midpoint (M) is calculated as:
Where
Lower class interval
Upper class interval
So, we have:
Class 63-65:
Class 66 - 68:
When the computation is completed, the frequency distribution will be:
Solving (b): Mean and standard deviation using 1-VarStats
Using 1-VarStats, the solution is:
<em>See attachment for result of 1-VarStats</em>
