Matthew's Responses in WyzAnt Answers

Let's use the method you listed above, but we need to correct the formula: (and note "T+a") is not the a that is in the first term, it is an angle, which I will represent with delta, δ):

a cos θ + b sin θ = c cos ( θ - δ )

The original equation, if we switch the order of the first two terms, will help us find a, b and c.

4 cos θ + 2 sin θ = sqrt(5)

From the above equation, a=4, b=2, and c=sqrt(a²+b²)=(sqrt(4²+2²))= sqrt(16+4)=sqrt(20).

Substituting these values into the method:

4 cos θ + 2 sin θ = sqrt(20) * cos ( θ - δ ) = sqrt(5)

The last equality, sqrt(5), comes from the original equation which looks just like the method on the left hand side of the equation.

Dividing both sides by sqrt(20):

cos ( θ - δ ) = sqrt(5) / sqrt(20) = sqrt(5/20) = sqrt(1/4) = 1/2.

So now we have cos ( θ - δ ) = 1/2.  Taking the inverse cosine on both sides:

θ - δ = cos^-1(0.5). 

Adding delta to both sides:

θ = cos^-1(0.5) + δ.

We need to determine δ to solve this equation. 

A handy relationship is that tan δ = b/a, so δ = tan^-1(b/a). 

b=2 and a=4, so δ = tan^-1(1/2) = tan^-1(0.5).

The phase shift δ in the cosine curve cos ( θ - δ ) is tan^-1(0.5), which is ~.464 radians or ~26.56 degrees.

So θ = cos^-1(0.5) + 26.56.  So for every angle where cos^-1(angle)=0.5, that is our solution (provided it lies within the 0 to 360 degree range stipulated at the beginning of the problem.

We know that this happens when the angle is -60 and +60 degrees, and that pair repeats every 360 degrees.  Adding -60 to 26.56 gives us an angle out of range.  60 works, as does -60+360=300.  60+360=420 is out of range too.  So 60 and 300 are valid angles, which result in theta values of:

θ = 60 + 26.56 = 86.56 degrees and θ = 300 + 26.56 = 326.56 degrees.

The curve will be a standard cosine curve shifted to the right 26.56 degrees and multiplied by sqrt(20), such that its range will not be +1 to -1 but +(sqrt(20)) to -(sqrt(20))... a 'taller' curve if you will by almost 4.5 times.

I let the client decide.  If the parent is comfortable with me coming to their house, I will.  My biggest concern is whether or not distractions at the home, such as siblings, electronic distractions, etc., will complicate the instruction.  I prefer "neutral ground" like a public library, given a choice, where it is quiet and distraction-free.

If you'll be using a textbook for class, perhaps you can buy it in advance, or gain access to the online textbook so you can begin getting familiar with the concepts in algebra.  That might help lessen your anxiety.  Plus remember you're going into the class to learn; no one expects you to understand the concepts in advance.  I know after 25 years starting college might be stressful, but, as an instructor (and a class of 1988 grad too I might add), we're there to help.  Don't be afraid to speak up when there's something you don't understand.  Best of luck to you! 

You need to convert this problem to a standard normal distribution to calculate the probability.  That means you convert the systolic number 120 to a Z-score.  Z = ( X - μ ) / σ, and you know all three variables.

Z = (120 - 120)/8 = 0/8 = 0.

P(X > 120) = P(Z > 0) = 0.5000, which can easily be looked up on a standard normal distribution probability table, or hopefully you remember that the normal curve is symmetric around Z=0 such that half of the area under the curve falls on either side of Z=0.

You could do it either way, but I'd learn towards n=48 for all questions since you are repeatedly referring to the 'fast track group'.  There should certainly be an expectation that not all candidates' data can be used; in the 'real world' you'll be tossing data if it is unusable.