Mellie

Oh right, this is only for the first. Thanks Alan:

Since we can add expected values of independent events, the expected value of the sum is the sum of the expected values of each.

Since we are just as likely to pick x as 1-x, the expected value of x (and likewise y) is equal to 1/2.

Therefore the expected value of the sum is 1/2+1/2=1.

A point  is chosen at random in the interior of equilateral triangle . What is the probability that  has a greater area than each of  and ?

I got it, it is 1/13. i lost the solution.

What do you mean by the end? We're you supposed to give us the money you had prior to buying the apples? If not, I will just calculate the amount you spend on the five apples. It's just multiplication. 

$${\mathtt{6.5}}{\mathtt{\,\times\,}}{\mathtt{5}} = {\frac{{\mathtt{65}}}{{\mathtt{2}}}} = {\mathtt{32.5}}$$

£32.50

Basically, the measures are like different variables because they have different values. So, first we want to add the meters.

2m+5m-3m=4m

Then, do the centimeters.

40cm+25cm-80cm=-15cm. 

You will get 3.85 m in total

$${\mathtt{207\,000\,600}}{\mathtt{\,\small\textbf+\,}}{\mathtt{207\,000\,600}} = {\mathtt{414\,001\,200}}$$

$${{\mathtt{13}}}^{{\mathtt{123}}} = {\mathtt{103\,521\,923\,711\,427\,330\,139\,508\,660\,561\,111\,399\,310\,356\,231\,562\,909\,825\,053\,544\,787\,423\,528\,887\,795\,603\,512\,927\,942\,762\,680\,653\,876\,132\,399\,975\,884\,814\,583\,882\,073\,552\,996\,857\,463\,797}}$$

Haha. I feel you

(Continued from CPhill)

98/2=49

So, you have 49s-3=49

Add 3 to each side.

49s=51

Divide both sides by 49

s=51/49... Not sure if this is right. Was there an x?

So, there is no x?