Answer:
Point E is given by 
Step-by-step explanation:
Let points C and D be
and let point E divides it in ratio
.
Then coordinates of point E are given by 
Put
Point E is given by 
Now plot point E.
Answer:
24
Step-by-step explanation:
4.8×5 you have 5 candy bars that are $4.80 each so you multiply that by 5.
Answer:
The difference would be 1/12.
Step-by-step explanation:
2/3 is 8/12.
Answer:
The answer is c) 761.0
Step-by-step explanation:
Mathematical hope (also known as hope, expected value, population means or simply means) expresses the average value of a random phenomenon and is denoted as E (x). Hope is the sum of the product of the probability of each event by the value of that event. It is then defined as shown in the image, Where x is the value of the event, P the probability of its occurrence, "i" the period in which said event occurs and N the total number of periods or observations.
The variance of a random variable provides an idea of the dispersion of the random variable with respect to its hope. It is then defined as shown in the image.
Then you first calculate E [x] and E [
], and then be able to calculate the variance.
![E[x]=0*\frac{1}{40} +10*\frac{1}{20} +50*\frac{1}{10} +100*\frac{33}{40}](https://tex.z-dn.net/?f=E%5Bx%5D%3D0%2A%5Cfrac%7B1%7D%7B40%7D%20%2B10%2A%5Cfrac%7B1%7D%7B20%7D%20%2B50%2A%5Cfrac%7B1%7D%7B10%7D%20%2B100%2A%5Cfrac%7B33%7D%7B40%7D)
![E[x]=0+\frac{1}{2} +5+\frac{165}{2}](https://tex.z-dn.net/?f=E%5Bx%5D%3D0%2B%5Cfrac%7B1%7D%7B2%7D%20%2B5%2B%5Cfrac%7B165%7D%7B2%7D)
E[X]=88
So <em>E[X]²=88²=7744</em>
On the other hand
![E[x^{2} ]=0^{2} *\frac{1}{40} +10^{2} *\frac{1}{20} +50^{2} *\frac{1}{10} +100^{2} *\frac{33}{40}](https://tex.z-dn.net/?f=E%5Bx%5E%7B2%7D%20%5D%3D0%5E%7B2%7D%20%2A%5Cfrac%7B1%7D%7B40%7D%20%2B10%5E%7B2%7D%20%2A%5Cfrac%7B1%7D%7B20%7D%20%2B50%5E%7B2%7D%20%2A%5Cfrac%7B1%7D%7B10%7D%20%2B100%5E%7B2%7D%20%2A%5Cfrac%7B33%7D%7B40%7D)
E[x²]=0+5+250+8250
<em>E[x²]=8505
</em>
Then the variance will be:
Var[x]=8505-7744
<u><em>Var[x]=761
</em></u>
Answer:
<h2>|||)</h2><h2>2^ (p + q) = 2^p • 2^q because you'd add the exponents, so xy</h2><h2>2^ 2q = 2^(q + q) = 2^q • 2^q as above, so y^2</h2><h2>2^ (p-1) = 2^p / 2^1 because you subtract exponents when dividing, so x/2</h2>
<h2>If xy = 32 then x = 32/y</h2><h2>so 2xy^2 would = 2(32/y)•y^2 = 64y = 32 so y = 1/2</h2><h2>making x = 64 since xy = 32</h2>
<h2>so 64 = 2^p; p = 6</h2><h2>1/2 = 2^q; q = -1</h2>