C.264.4 that should answer your question
Answer:
Step-by-step explanation:
Given two upward facing parabolas with equations
The two intersect at
=
x=
area enclosed by them is given by
A=![\int_{-\sqrt{\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left [ \left ( x^2+2\right )-\left ( 6x^2\right ) \right ]dx](https://tex.z-dn.net/?f=%5Cint_%7B-%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%7D%5E%7B%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%7D%5Cleft%20%5B%20%5Cleft%20%28%20x%5E2%2B2%5Cright%20%29-%5Cleft%20%28%206x%5E2%5Cright%20%29%20%5Cright%20%5Ddx)
A=
A=![4\left [ \sqrt{\frac{2}{5}} \right ]-\frac{5}{3}\left [ \left ( \frac{2}{5}\right )^\frac{3}{2}-\left ( -\frac{2}{5}\right )^\frac{3}{2} \right ]](https://tex.z-dn.net/?f=4%5Cleft%20%5B%20%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%20%5Cright%20%5D-%5Cfrac%7B5%7D%7B3%7D%5Cleft%20%5B%20%5Cleft%20%28%20%5Cfrac%7B2%7D%7B5%7D%5Cright%20%29%5E%5Cfrac%7B3%7D%7B2%7D-%5Cleft%20%28%20-%5Cfrac%7B2%7D%7B5%7D%5Cright%20%29%5E%5Cfrac%7B3%7D%7B2%7D%20%5Cright%20%5D)
A=
<span>Exactly 33/532, or about 6.2%
This is a conditional probability, So what we're looking for is the probability of 2 gumballs being selected both being red. So let's pick the first gumball.
There is a total of 50+150+100+100 = 400 gumballs in the machine. Of them, 100 of the gumballs are red. So there's a 100/400 = 1/4 probability of the 1st gumball selected being red.
Now there's only 399 gumballs in the machine and the probability of selecting another red one is 99/399 = 33/133.
So the combined probability of both of the 1st 2 gumballs being red is
1/4 * 33/133 = 33/532, or about 0.062030075 = 6.2%</span>
Answer:
The answer is B ( X>6 )
Step-by-step explanation:
2x+4= 16
2x = 16 - 4
2x = 12
x = 12/2
x= 6
x>6
Answer:
and 
Step-by-step explanation:
Coefficients are simply the numerical constants "attatched" (being multiplied by) [to] the variables.
