GYanggg

Hey Guest! 

This is a basic trigonometrey relationship. 

We use cosine in this case, where cosine is adjacent over hypotenuse. 

\(\cos{30º}=\frac{x}{10}\)

\(\frac{\sqrt3}{2}=\frac{x}{10}\)

\(x=\boxed{5\sqrt3}\)

I hope this helped,

Gavin

To see if it is possible, we can just say you aced your final, 100% and see if it exceeds 66.5%.

Your current grade is 66.5% and it represents 90% of your final grade. 

If you aced your final, 100%, it will still represent 10% of your final grade. 

\({0.665\cdot0.9+1\cdot0.1}=0.6985\)

You will get a final grade of 69.85% even if you aced your final. 

It depends what your teacher rounds up by. 

If he/she rounds up by whole number, you will get 70%.

If he/she rounds up by tenth, you will get 69.9%, oof.

If she doesn't round, you will get 69.85%. 

I hope this helped,

Gavin

1. All we need to do is find the expected value for one die, then do what the problem asks us to do. 

The expected value for the first die was:

\(\frac{1+2+3+4+5+6}{6}=3.5\), 

\(3.5+3.5=\boxed7\),

That is the expected value for the outcome.

2.

The probabilty of getting a sum of 2 is: \(\frac{1}{36}\). The numerator is one, because there is one way of getting a sum of 2, 1 + 1. 

We do the same for the rest of the sums, 

The probabilty of getting a sum of 3 is: \(\frac{2}{36}\), (1,2) ; (2,1)

The probabilty of getting a sum of 4 is: \(\frac{3}{36}\), (1,3) ; (3,1) ; (2,2)

The probabilty of getting a sum of 5 is: \(\frac{4}{36}\), (1,4) ; (4,1) ; (2,3) ; (3,2)

The probabilty of getting a sum of 6 is: \(\frac{5}{36}\), (1,5) ; (5,1) ; (2,4) ; (4,2) ; (3,3) 

The probabilty of getting a sum of 7 is: \(\frac{6}{36}\), (1,6) ; (6,1) ; (2,5) ; (5,2) ; (3,4) ; (4,3)

The probabilty of getting a sum of 8 is: \(\frac{5}{36}\), (2,6) ; (6,2) ; (3,5) ; (5,3) ; (4,4)

The probabilty of getting a sum of 9 is: \(\frac{4}{36}\), (3,6) ; (6,3) ; (4,5) ; (5,4) 

The probabilty of getting a sum of 10 is: \(\frac{3}{36}\), (4,6) ; (6,4) ; (5,5) 

The probabilty of getting a sum of 11 is: \(\frac{2}{36}\), (5,6) ; (6,5) 

The probabilty of getting a sum of 12 is: \(\frac{1}{36}\), (6,6)  

The most likely outcome is getting a 7, with a 1/6 probability. 

3. 

Greatest is 6 + 6 = 12

Smallest is 1 + 1 = 2

Hey Guest!

The smallest is with the two smallest numbers on the die, 1 * 1 = 1.

The largest is with the two largest numbers on the die, 6 * 6 = 36. 

All we need to do is find the expected value for one die, then do what the problem asks us to do. 

\(\frac{1+2+3+4+5+6}{6}=3.5\), 

We already got the expected value for one die, the product is just:

\(3.5\cdot3.5=\boxed{12.25}\)

I hope this helped,

Gavin

2. 

All we need to do is find the expected value for one die, then do what the problem asks us to do. 

The same with the first problem, the expected value for the first die was:

\(\frac{1+2+3+4+5+6}{6}=3.5\), 

\(3.5+3.5=\boxed7\), which was what we got after all that work. 

This method is way easier and harder to make careless mistakes. 

We can do the same with #2. We already got the expected value for one die, the product is just:

\(3.5\cdot3.5=\boxed{12.25}\)

I hope this helped,

Gavin

Hey Guest! I'm not too familiar with expected value, but I can still give it a go:

NOTE: THIS SOLUTION IS WAY TO DIFICULT, LOOK AT MY SOLUTION TO 2. FOR THE CORRECT ONE. 

I'm not sure if this is the best way to do this, but it works. 

The possible sums of two dice are \((2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)\)

The probabilty of getting a sum of 2 is: \(\frac{1}{36}\). The numerator is one, because there is one way of getting a sum of 2, 1 + 1. 

We do the same for the rest of the sums, 

The probabilty of getting a sum of 3 is: \(\frac{2}{36}\), (1,2) ; (2,1)

The probabilty of getting a sum of 4 is: \(\frac{3}{36}\), (1,3) ; (3,1) ; (2,2)

The probabilty of getting a sum of 5 is: \(\frac{4}{36}\), (1,4) ; (4,1) ; (2,3) ; (3,2)

The probabilty of getting a sum of 6 is: \(\frac{5}{36}\), (1,5) ; (5,1) ; (2,4) ; (4,2) ; (3,3) 

The probabilty of getting a sum of 7 is: \(\frac{6}{36}\), (1,6) ; (6,1) ; (2,5) ; (5,2) ; (3,4) ; (4,3)

The probabilty of getting a sum of 8 is: \(\frac{5}{36}\), (2,6) ; (6,2) ; (3,5) ; (5,3) ; (4,4)

The probabilty of getting a sum of 9 is: \(\frac{4}{36}\), (3,6) ; (6,3) ; (4,5) ; (5,4) 

The probabilty of getting a sum of 10 is: \(\frac{3}{36}\), (4,6) ; (6,4) ; (5,5) 

The probabilty of getting a sum of 11 is: \(\frac{2}{36}\), (5,6) ; (6,5) 

The probabilty of getting a sum of 12 is: \(\frac{1}{36}\), (6,6)  

The expected value is: 

\((\frac{1}{36}\cdot2)+(\frac{2}{36}\cdot3)+(\frac{3}{36}\cdot4)+(\frac{4}{36}\cdot5)+(\frac{5}{36}\cdot6)+(\frac{6}{36}\cdot7)+(\frac{5}{36}\cdot8)+(\frac{4}{36}\cdot9)+(\frac{3}{36}\cdot10)+(\frac{2}{36}\cdot11)+(\frac{1}{36}\cdot12)\)

When you simplify: 

\(\frac1{18}+\frac16+\frac13+\frac59+\frac56+\frac76+\frac{10}{9}+\frac11+\frac{10}{12}+\frac{11}{18}+\frac13\)

When you add these up, you get 7. 

I hope this helped,

Gavin

Hey Guest!

Here is an ordinary polar graph:

It is like an ordinary x-y coordinate graph, but the x is wrapped around in a cicle. 

You can first graph the x or the y, it doesn't matter. 

For ( 5, 240°), let's just say we first graph 5. If you look at the arrow, there is a 5 between the 4 and 6. 

It is the degrees that determines the specific point. 

Then, we find the 240° line, then find the intersection of the 5 circle and the 240° line. 

You can try (-1, 120°) on your own. 

I hope this helped,

Gavin.

Hey QS!

The problem already gives you function A and B, you just need to write an equation. 

Function A: \((x + 4)\cdot2=y\)

Function B: \(x\div5+1=y\)

M goes through A and then B to turn into 2p+1

We write the equation:

\([(m+4)\cdot2]\div5+1=2p+1\\ (2m+8)=2p\cdot5\\ m+4=5p\\ m=5p-4\\ \)

I hope this helped,

Gavin

b) 

In mathematics, an inverse function (or anti-function) is a functionthat "reverses" another function.

The function the problem gives us is basically: y = x/5 + 1

To inverse it, we write it in the form, x = ...

First we subtract 1 from both sides, to get y - 1 = x/5

Then we multiply both sides by 5, we get: 5y - 5 = x 

It becomes, x = 5y - 5, or x = 5(y-1) 

So the function is subtract one, then multiply the difference by 5. 

I hope this helped,

Gavin

a) 

As the rule states, we first add 4, then multiply the sum by 2. 

The first imput gives us 6, we follow the fuction's rules:

6 + 4 = 10, 10 * 2 = 20 

The output, or y, is 20. 

The second part, gives us the output. We can just reverse it. 

What times 2 is 13? 13/2 = 6.5.

What plus 4 is 6.5? 6.5 - 4 = 2.5.

The imput is 2.5. 

The first blank is 20. 

The second blank is 2.5. 

I hope this helped,

Gavin