<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>
<span>a/b = 3/5 ----- a = (3b)/5
a + b = 136 --- (3b)/5 + b = 136
3b + 5b = 680
8b = 680
b = 85
a = (3b)/5 = (3 * 85)/5 = 51
a = 85 and b = 51</span>
I actually got zero for this when I did this in phothomath
Hello,
P(x)=x^4-6x²+2=(x-a)(x-b)(x-c)(x-d)
=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2-(abc+abd+acd+bcd)x+abcd
==>
ab+ac+ad+bc+bd+cd=-6
abc+abd+acd+bcd=0
abcd=2
a+b+c+d=0 ==>(a+b+c+d)²=0=a²+b²+c²+d²+2(ab+ac+ac+bc+bd+cd)
==>a²+b²+c²+d²=0-2*(-6)=12
if a is a root P(a)=0==>a^4-6a²+2=0
if b is a root P(b)=0==>b^4-6b²+2=0
if c is a root P(a)=0==>c^4-6c²+2=0
if d is a root P(a)=0==>d^4-6d²+2=0
==>a^4+b^4+c^4+d^4-6(a²+b²+c²+d²)+4*2=0
==>a^4+b^4+c^4+d^4=-8+6*12=64
