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I don't know much about Sets or Subsets, but it seems to me you may have overcounted the numbers in the Set, especially when he/she says "Two different numbers are selected simultaneously and at random from the set", which sounds to me that a pairing of 2 numbers, say {2,5} is the same as{5,2}. In other words, you have:

7 nCr 2 = 21 distinct ways  of choosing your numbers as oulined below:

{1, 2} | {1, 3} | {1, 4} | {1, 5} | {1, 6} | {1, 7} | {2, 3} | {2, 4} | {2, 5} | {2, 6} | {2, 7} | {3, 4} | {3, 5} | {3, 6} | {3, 7} | {4, 5} | {4, 6} | {4, 7} | {5, 6} | {5, 7} | {6, 7} (total: 21). And the differences between them would be:

1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6 >> for a total of 21. 15 of them have a difference of 2 and >.

Then, according tto this scenario, the probability would be: 15 / 21 = 5/7

Note: Somebody more versed in Set Theory should take a look at this question. Thanks.

a is the slope and b the y intercept. We solve for the slope first. Plugging the points into the slope formula, we find that a is 6/4=3/2. So now we have y=3/2x+b. Then we plug in either of the given points to solve for b. 

y=3/2x+b

5=3/2(4)+b

5=6+b

b=-1

Therefore the equation of the line is y=3/2x-1.

\(\frac{1}{3}(6z+18)=3z-9\\ 2z+6=3z-9\\ 6+9=3z-2z\\ \boxed{z=15}\)

The value of z = 15

I am so sorry it is incorrect.

no,

(where I can get any number starting from x)

You can use the state diagram approach for these kinds of problems.  This gives you (e_1,e_2) = (2,4).

Reference: http://www.aquatutoring.org/ExpectedValueMarkovChains.pdf

(x,y,z) = (8/5,8/3,-8)

Mike is riding his bike at 15 miles perhour.  What is his speed in feet per second?  (Use 5280 feet in each mile) 

You already have your mathematically computed answer; I'm just going to tell you a way to get an estimate. 

When you have miles per hour, multiply it by 1.5 to get feet per second.  In this problem the estimate is 15 x 1.5 = 22.5. 

The accurate answer geno gave you was 22 , this estimate method gives you 22.5.  Close enough for government work. 

_.

Assume a bag of wheat costs =$1.00 per pound
Then a bag of rice would cost  =$1.20 per pound
1.00 - [1.00/1.20] = 16 2/3 % - This is how much cheaper wheat is per pound. [Units don't matter, whether they are in pounds, kg, tons....etc.]
This means that if 1 pound of rice cost $1, then the cost of 1 pound of wheat would be: $1 - [0.1667 * $1] =0.8333 cents.

My account is robot317 

the link is here:

For example, I have the numbers 3 and 5 I know that I can get any number starting from 8 by adding diffrent combinations of these numbers ie 9 = 3+3+3 and 10 = 5+5. How, knowing that the numbers guaranteedly have this point (where I can get any number starting from x), how could I estimate what x is?

Two different numbers are selected simultaneously and at random from the set {1, 2, 3, 4, 5, 6, 7. What is the probability that the positive difference between the two numbers is 2 or greater? Express your answer as a common fraction.  

I can grasp just from thinking that any two numbers will work if they're not consecutive.  That is, 1,2 won't work, nor 2,3, nor 3,4, etc.  

I don't know how to figure probabilities, so I have to make a grid and count them.  In the grid below, the colored pairs won't work. 

1,2   1,3   1,4   1,5   1,6   1,7

2,1   2,3   2,4   2,5   2,6   2,7

3,1   3,2   3,4   3,5   3,6   3,7

4,1   4,2   4,3   4,5   4,6   4,7

5,1   5,2   5,3   5,4   5,6   5,7

6,1   6,2   6,3   6,4   6,5   6,7

There are 36 pairs, and 11 of them won't work.  That means 25 of them will work.  So your probability is 25/36.

_.

Mine is robot317 

https://artofproblemsolving.com/community/user/543357

Yep I am

thank you!

Here's the answer according to your textbook: 

240 / 26 = 9.230769231  

https://web2.0calc.com/questions/help-5_28

\(\begin{array}{|rcll|} \hline \tan(A) &=& \dfrac{3}{2} \\\\ \tan(A) &=& \dfrac{x}{p} \\\\ \hline \dfrac{3}{2} &=& \dfrac{x}{p} \\ \dfrac{2}{3} &=& \dfrac{p}{x} \\ p &=& \dfrac{2}{3}*x \\ \hline \end{array} \begin{array}{|rcll|} \hline \tan(C) &=& \dfrac{2}{3} \\\\ \tan(C) &=& \dfrac{x}{q} \\\\ \hline \dfrac{2}{3} &=& \dfrac{x}{q} \\ \dfrac{3}{2} &=& \dfrac{q}{x} \\ q &=& \dfrac{3}{2}*x \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline p + x + q &=& AC \\\\ \dfrac{2}{3}*x + x+ \dfrac{3}{2}*x &=& AC \quad | \quad AC = \sqrt{13} \\\\ \dfrac{2}{3}*x + x+ \dfrac{3}{2}*x &=& \sqrt{13} \\\\ x \left( \dfrac{2}{3} +1+ \dfrac{3}{2} \right) &=& \sqrt{13} \\\\ x \left( \dfrac{4}{6} +\dfrac{6}{6}+ \dfrac{9}{6} \right) &=& \sqrt{13} \\\\ x * \dfrac{19}{6} &=& \sqrt{13} \\ \mathbf{x} &=& \mathbf{\dfrac{6}{19}* \sqrt{13}} \\ \hline \end{array}\)

The side length of the square is \(\mathbf{\dfrac{6}{19}* \sqrt{13}}\)

https://web2.0calc.com/questions/help-3_37

\(\begin{array}{|rcll|} \hline 2*\text{area}_{PQR} &=& 12*12*\sin(120^\circ) \quad | \quad \sin(120^\circ)= \sin(60^\circ)=\dfrac{\sqrt{3}} {2} \\ 2*\text{area}_{PQR} &=& 12*12*\dfrac{\sqrt{3}} {2} \\ 2*\text{area}_{PQR} &=& 12*6* \sqrt{3} \\ \text{area}_{PQR} &=& 6*6* \sqrt{3} \\ \mathbf{\text{area}_{PQR}} &=& 36* \mathbf{\sqrt{3}} \\ \hline \end{array}\)

https://web2.0calc.com/questions/help-2_42

\(\begin{array}{|rcll|} \hline 3CD &=& 12 \\ CD &=& \dfrac{12}{3} \\ \mathbf{CD} &=& \mathbf{4} \\ \hline AD &=& 3CD \\ AD &=& 3*4 \\ \mathbf{AD} &=& \mathbf{12} \\ \hline AC &=& AD+CD \\ AC &=& 12 + 4 \\ \mathbf{AC} &=& \mathbf{16} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline AB^2 + BC^2 &=& AC^2 \quad | \quad \mathbf{AC=16} \\ AB^2 + BC^2 &=& 16^2 \quad | \quad AB^2 = 12^2+BD^2 \\ 12^2+BD^2 + BC^2 &=& 16^2 \quad | \quad BC^2 = 4^2+BD^2 \\ 12^2+BD^2 + 4^2+BD^2 &=& 16^2 \\ 2BD^2 &=& 16^2 -12^2-4^2 \\ 2BD^2 &=& 256-144-16 \\ 2BD^2 &=& 96 \\ BD^2 &=& 48 \\ BD^2 &=& 16*3 \\ \mathbf{BD} &=& \mathbf{4\sqrt{3}} \\ \hline \end{array}\)

https://web2.0calc.com/questions/help-1_39

\(\begin{array}{|rcll|} \hline \cos(60^\circ) &=& \dfrac{CB}{12} \\ CB &=& 12*\cos(60^\circ) \quad | \quad \cos(60^\circ) = \dfrac12 \\ CB &=& 12*\dfrac12 \\ \mathbf{CB} &=& \mathbf{6} \\ \hline \tan{30^\circ} &=& \dfrac{XC}{CB} \\\\ \tan{30^\circ} &=& \dfrac{XC}{6} \\ XC &=& 6*\tan{30^\circ} \quad | \quad \tan{30^\circ} = \dfrac{\sin(30^\circ)} {\cos(30^\circ)}= \dfrac{\dfrac{1}{2}} {\dfrac{\sqrt{3}} {2} } = \dfrac{1}{\sqrt{3}} \\ XC &=& 6*\dfrac{1}{\sqrt{3}} \\ XC &=& 6*\dfrac{1}{\sqrt{3}}*\dfrac{\sqrt{3}}{\sqrt{3}} \\ \mathbf{XC} &=& \mathbf{2* \sqrt{3}} \\ \hline 2*\text{area}_{BXA} &=& 12*h \quad | \quad h = XC \\ 2*\text{area}_{BXA} &=& 12*XC \quad | \quad \mathbf{XC=2* \sqrt{3}} \\ 2*\text{area}_{BXA} &=& 12*2* \sqrt{3} \\ \mathbf{\text{area}_{BXA}} &=& \mathbf{12*\sqrt{3}} \\ \hline \end{array}\)