Integration Area problem

Q4. find the region bounded by the two curves.

On QUORA

Mapping domain of composite functions question

f:R+ maps to R, f(x) = x to the power of -0.5, and g: R maps to R. g(x)=3-x.
a Show f(g(x)) is not defined
b restrict the domain to make a well defined function.

On QUORA

Variance Question, year 12 Specialist in Victoria

Notes

The Question

Specifically

Specifically, the student has been given E(x) and tasked to find Var(5–3x). Their notes don’t actually cover it, neither does the textbook address it well.

A biased die with six sides is rolled. The discrete random variable X represents the score on the uppermost face. The probability distribution is shown in the table
c) Given E(x) = 4.2 find a and b
d) Show E(x squared) = 20.4
e) find Var(5–3x)
===
From notes, Var (x) = E(x squared) - E((x) squared)
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Solution via Jan van Delden, MSc Math and still interested

Yes I can, now that I tried to read this elaborate link.

The point is that once you know var(𝑥)$\mathrm{var}(x)$ you should be able to apply a few basic rules of the variance operator in order to compute var(5–3𝑥)$\mathrm{var}(5–3x)$.

We have, with 𝑎$a$ a constant:

• var(𝑥+𝑎)=var(𝑥)$\mathrm{var}(x+a)=\mathrm{var}(x)$
• var(𝑎𝑥)=𝑎2var(𝑥)$\mathrm{var}(ax)=a^2\mathrm{var}(x)$

But first you should compute var(𝑥)$\mathrm{var}(x)$ using the equation, which is actually given to you in the link.

EKC Hallam Q2 solved

Solution, sum each column to a square. Answer c

EKC Hallam Triplet Sequence

Is it the first term times 16 minus the last? Cf Quora

That would suggest the answer is -44, and so e.
===
Seth Wax suggested this better answer

The second number in each triple is the sum of the cubes of the first and the third.

43+13=64+1=65$4^3+1^3=64+1=65$

23+33=8+27=35$2^3+3^3=8+27=35$

(−3)3+43=−27+64=37

Answer is C

Arrangements with Mr Mak

Q In how many ways can 4 boys and three girls be arrange in a straight line, If there are 8 boys and five girls to choose from?


AM 8C4 = 70 ways of choosing the boys5C3 = 10 ways of choosing the girls
Then 7! = 5040 ways of arranging the chosen boys and girls

Giving a total of 70 x 10 x 5040 = 3528000 ways.

Me I thought that and didn’t feel confident about the answer .. would there be limit beneath the seven factorial for genericboys or girls being in similar position?

AM No, only if they're identical boys / identical girls. If unconvinced, treat the boys as B1, B2, B3, B4, and the girls as G1, G2 and G3 to see that you are arranging 7 unique elements.

Resolving forces with Mr Mak Q 11 Spec Math

Question 11
A string is connected to two points A and D in a horizontal line and masses of 12kg and W kg are attached at points B and C. If AB, BC and CD make angles of 40°, 20°, and 50° respectively with the horizontal, calculate the tensions in the string and the value of W.

 I think they use kg Wt in the textbook, which means you don’t have to use g (gravity).

An Ant walk with Alvin Mak answer

An ant walks 8 steps on a number plane, starting at the origin and finishing there. How many different paths are available?



To start at (0,0) and finish at (0,0), the ant must have the same number of "north" steps as "south" steps, likewise with the "east" and "west" steps. Let N, S, E, W represent these directional steps, the problem reduces to how many ways to arrange a chain of 8 such letters such that there's equal number of N's and S's, and equal number of E's and W's, possibilities are:

1) 4N's and 4S's: 8!/(4!4!) = 70
2) 3N's, 3S's, 1E and 1W: 8!/(3!3!) = 1120
3) 2N's, 2S's, 2E's and 2W's: 8!/(2!2!2!2!) = 2520
4) 1N, 1S, 3E's and 3W's: 8!/(3!3!) =1120
5) 4E's and 4W's: 8!/(4!4!) = 70

Totalling to 4900