I dont understand what you are asking
<span>Winning Probablity = 0.2, hence Losing Probability = 0.8
Probablity of winning atmost one time, that means win one and lose four times or lose all the times. So p(W1 or W0) = p (W1) + p(W0)
Winning once W1 is equal to L4, winning zero times is losing 5 times.
p(W1) = p(W1&L4) and this happens 5 times; p(W0) = p(L5);
p (W1) + p(W0) = p(L4) + p(L5)
p(L4) + p(L5) = (5 x 0.2 x 0.8^4) + (0.8^5) => 0.8^4 + 0.8^5
p(W1 or W0) = 0.4096 + 0.32768 = 0.7373</span>
Answer:
Possible C:
(1 , 1 + 3sqrt(3))
(1 , 1 - 3sqrt(3))
Step-by-step explanation:
AB = 6 units
Midpoint of AB is (1,1)
C would be at x = 1
So, (1,y)
(y-1)² + (1+2)² = 6²
(y-1)² = 36 - 9
(y-1)² = 27
y - 1 = 3sqrt(3)
y - 1 = -3sqrt(3)
y = 1 + 3sqrt(3)
y = 1 - 3sqrt(3)
Answer:
41
Step-by-step explanation:
So basically you need to use the pythagorean thrm. Essentially, the diagonal and the width and length create a triangle, the diagonal being the hypothenuse. So the formula goes as ![\sqrt[]{a^{2} +b^{2} }](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7Ba%5E%7B2%7D%20%2Bb%5E%7B2%7D%20%7D)
. the variable "a" being the length and the variable "b" being the width. So plug it in and there you have your answer!
Answer:
volume = 338 x 20 = 6760 cm^3
Step-by-step explanation:
large rectangular top area = 14 x 22 = 308 cm^2
small rectangular top area = 6x5 = 30 cm^2
total top area = 308 + 30 = 338 cm^2
volume = 338 x 20 = 6760 cm^3
