y < - 8 or y > 4
inequalities of the form | x | > a always have solutions of the form
x < - a or x > a
we have to solve
y + 2 < - 6 or y + 2 > 6
y + 2 < - 6 ( subtract 2 from both sides )
y < - 8
or
y + 2 > 6 ( subtract 2 from both sides )
y > 4
these can be combined using interval notation
y ∈ (- ∞, - 8 ) ∪ (4, ∞ )
As a check
substitute chosen values of x from each interval
y = - 10 : | - 10 + 2 | = | - 8 | = 8 > 6 this is true
y = 12 : | 12 + 2 | = | 14 | = 14 > 6 which is also true
Answer: <span>A square inscribed in a circle.
</span>
Justification:
Note that by making two perpendicular lines that intersect each other in the center of the circle, he obtains 4 equidistant points on the circumference.
So, joining each pair of neighbouring points, the image will reveal 4 congruent sides joining at right angles (90°). This is the image of a square with the four vertices on the circumference.
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>
the answer is C) home loan
Use Photomath it works well