For simple integer factors as this one has, you want to find two values for the quadratic in the form ax^2+bx+c. Let the two values be j and k. These two values must satisfy two conditions.
jk=ac=10 and j+k=b=11, so j and k must be 1 and 10.
Now replace bx with jx and kx...
2x^2+x+10x+5 now factor 1st and 2nd pair of terms.
x(2x+1)+5(2x+1)
(x+5)(2x+1)
Answer:
x = infinite amount of solutions
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
Step-by-step explanation:
<u>Step 1: Define Equation</u>
8(2x + 5) = 16x + 40
<u>Step 2: Solve for </u><em><u>x</u></em>
- Distribute 8: 16x + 40 = 16x + 40
- Subtract 40 on both sides: 16x = 16x
- Divide 16 on both sides: x = x
Here we see that <em>x</em> does indeed equal <em>x</em>.
∴ <em>x</em> has an infinite amount of solutions.
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>
Because when you do 3 x 2=6 and then 52-6 it doesn’t equal 20. It equals 46.
Answer:
Exponents are simply repeated multiplication so the answer is 7 * 7 * 7 * 7.
