GYanggg

The probability that a girl is chosen is 2/3 the probability that a boy is chosen.

This means that if there are x boys, there are a total of 2/3 x girls. 

The total amount of kids in the class is 2/3x + 1x = 5/3x.

This is equal to 20. 

\(\frac53x=20\Rightarrow{x}=12\)

Therefore, there are \(12\cdot\frac23=8\) girls. 

We can verify this. The class has 20 people, 8 girls and 12 boys. 

The probabilty a girl is chosen is 8/20 and boy is 12/20. 

8/20 is also 2/3 of 12/20, so our answer is correct.

I hope this helped,

Gavin

First we need to figure out how many ways to draw two cards from a standard deck of 52 cards. 

To do this, we get: \(\binom{52}{2}=1326\)

The ways to chose them so their rank is the same is harder. 

There are 13 different ranks, and in each rank, their are 4 suits. 

To choose two cards in a rank, we have \(\binom42=6\) ways to do this. 

Since there are 13 ranks, we have: \(13\cdot6\) ways to chose the cards so their rank is the same. 

The probabilty is then \(\frac{78}{1326}=\frac1{17}\)

I hope this helped,

Gavin.

The sample mean is the average of all the numbers in the sample. 

\(\frac{25+23+26+29+37+23+32+29+25+22+31+34}{12}=\frac{336}{12}=\boxed{28}\)

Therefore, 28 is the sample mean.

The population mean is the average of the whole set. 

This is already given in the problem, 27.

I hope this helped,

Gavin.

The smallest integer that has 4 digits in base 8 is \(1000_8\), which stands for \(8^3 = 2^9\). The largest integer that has 4 digits in base 8 is \(7777_8\), which is 1 less than \(10000_8\) and therefore stands for \(8^4-1 = 2^{12}-1.\)

Thus, when a 4-digit base-8 integer is written in base 2, its highest place value is either \( 2^9, 2^{10},\) or \(2^{11}\). It follows that the base-2 expression has 10, 11, or 12 digits, so the sum of all possible values for d is \(10+11+12 = \boxed{33}.\)

I hope this helped,

Gavin

I already drew a diagram, but don't know how to upload it from my computer.

First you connect the point A to the two points that construct the line segment labeled 10. 

AD is congruent to the line segment perpendicular to the side labelled 10. 

You have constructed a 30 - 60 - 90 triangle. The hypotenuse is 20.

The leg is the square root of 20^2 - 10^2.

Can someone tell me how to upload an image, then this solution would be much clearer. 

Thanks

If:

 \(y=x+6\\ y=x^2+8x+16\)

then: \(x+6=x^2+8x+16\)

\(x^2+7x+10=0\\ (x+2)(x+5)=0\\ x_1=-2, x_2=-5 \)

If x = -2, y = 4

If x = - 5, y = 1

I hope this helped,

Gavin

I have noticed a flaw in my proof. 

One may not understand why \(x^3+y^3+z^3−3xyz=(x+y+z)(x^2+y^2+z^2−xy−xz−yz)=0\)

Here is why: 

Using \(a^3 + b^3 = (a + b)(a^2 – ab + b^2)\), 

We get: \(x^3+y^3+z^3−3xyz=x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz\)

Since I added and subtracted \(3x^y\ \text{and}\ 3xy^2\), the value of the expression does not change. 

I did this so I can factor out some values: 

\(\boxed{x^3+y^3+z^3−3xyz\\ =x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3x^2y-3xyz\\ =(x+y)^3+z^3-3xy(x+y+z)\\ =(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)\\ =(x+y+z)(x^2+y^2+z^2-xy-xz-yx)}\)

If you still don't understand something, private message me and I will try to resolve the issue. 

Gavin

Prove: \((a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)\)

Using the method your proofs say, here is what they mean. 

Basically, the proof is assigning each expression another varible. 

\(a-b=x\\ b-c=y\\ c-a=z\\\)

So instead of a - b, b - c, or c - a, they replace these expressions with x, y, and z. 

When we do replace them with x, y, and z, we get:

\(x^3+y^3+z^3=3(x)(y)(z)\)

Since \(x+y+z=a-b+b-c+c-a=0\), we get \(x + y +z = 0\)

Using that, we get:

\(x^3+y^3+z^3−3xyz=(x+y+z)(x^2+y^2+z^2−xy−xz−yz)=0\)

\(a-b=0\Rightarrow{a}=b\)

Therefore, \((a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)\)

If hope this helped,

Gavin.

Hey Guest!

We can set up the system of equations if we assign the cost of each item a variable:

Cost of pretzel is \(x\)

Cost of lemonade is \(y\)

\(10x-5y=40\Rightarrow50x-25y=200\rightarrow(1)\\ 25x-2y=115\Rightarrow50x-4y=230\rightarrow(2)\\\)

\((2)-(1)\\ 21y=30\\ y\approx1.43\\ x\approx4.71\)

Therefore the cost of lemonade is $4.71 and the cost of the preztels is $1.43

I hope this helped,

Gavin

Hey bobbly!

 \(\frac5{x+3}+\frac{3}{x-2}=4\\ \frac{5\cdot(x-2)+3\cdot(x+3)}{(x-2)(x+3)}=4\\ \frac{5x-10+3x+9}{x^2+x-6}=4\\ 5x-10+3x+9=4x^2+4x-24\\ 4x^2-4x-23=0\)

We then use the quadratic formula: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \), to solve for x. 

\(x_1=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\cdot \:4\left(-23\right)}}{2\cdot \:4}=\frac{1+2\sqrt{6}}{2}\)

\(x_2=\frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:4\left(-23\right)}}{2\cdot \:4}=\frac{1-2\sqrt{6}}{2}\)

I hope this helped,

Gavin