I got 7.1 x 10^(-11) but it's been a few years so I could've done it wrong
The vertex point is (0,4)
The two other points are (-2,0)(2,0)
ANSWER:
-78-a
= -78-(-29)
= -78+29
= -49
<span><span>n<span>x4/</span></span>5</span>=<span>3/<span>4
</span></span><span><span><span><span>1/5</span><span>n<span>x^4</span></span></span><span><span>x^4/</span>5</span></span>=<span><span>3/4</span><span><span>x^4/</span>5</span></span></span><span>
Answer is n=<span>15/<span>4<span>x<span>4</span></span></span></span></span>
Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
P(brown) = 12% = 0.12
P(Yellow) = 15% = 0.15
P(Red) = 12% = 0.12
P(blue) = 23% = 0.23
P(orange) = 23% = 0.23
P(green) = 15% = 0.15
A.) Compute the probability that a randomly selected peanut M&M is not yellow.
P(not yellow) = P(Yellow)' = 1 - P(Yellow) = 1 - 0.15 = 0.85
B.) Compute the probability that a randomly selected peanut M&M is brown or red.
P(Brown) or P(Red) :
0.12 + 0.12 = 0.24
C.) Compute the probability that three randomly selected peanut M&M’s are all brown.
P(brown) * P(brown) * P(brown)
0.12 * 0.12 * 0.12 =0.001728
D.) If you randomly select three peanut M&M’s, compute that probability that none of them are blue.
P(3 blue)' = 1 - P(3 blue)
P(3 blue) = 0.23 * 0.23 * 0.23 = 0.012167
1 - P(3 blue) = 1 - 0.012167 = 0.987833
If you randomly select three peanut M&M’s, compute that probability that at least one of them is blue.
P(1 blue) + p(2 blue) + p(3 blue)
(0.23) + (0.23*0.23) + (0.23*0.23*0.23)
0.23 + 0.0529 + 0.012167
= 0.295067
