Answer:
543
Step-by-step explanation:
Answer:
11/14
Step-by-step explanation:
(P of blue marble) + (P of even numbered) - (P blue and even) =
8/14 + 7/14 - 4/14 = 11/14
Answer:
17. z=18
18. n=-15
19. j = -50
20. a=-27
21. h=-15
22. d=.392
Step-by-step explanation:
17. Multiply both sides by 3, then subtract 6 from both sides
18. Multiply both sides of the equation by 2, then add 7
19. Multiply both sides by -4, then subtract 18 from both sides.
20. Add six to both sides, then multiply both sides by 3
21. Subtract 4 from both sides, then multiply both sides by 4
22. Subtract 6.42 from both sides, then divide both sides by -10
There are 36 outcomes when rolling a pair of dice.
The ones adding up to 8 are: (2,6),(3,5),(4,4),(5,3),(6,2).
So the probability of rolling a sum of 8 is 5/36.
With three (independent) random throws, the probability of having a sum of 8 at least once equals 1-probability of no 8 at all.
P(not 8)=1-5/36=31/36
For this to happen three times, we use the multiplication rule
P(not 8 three times)=(31/36)^3=29791/46656=
Therefore
P(sum of 8 at least once)=1-29791/46656=16865/46656
[=0.3615 approx.]
Step-by-step explanation:
a. The random variable is a binomial distribution.
b. the sample space, X = {0, 1, 2, 3, 4, 5}
the pmf
we solve for this using
nCx * P^x * (1-p)^n-x
n = 5
p = 0.3
for x = 0
5C0 * 0.3⁰(1-0.3)^5-0
= 1 * 1* 0.7⁵
= 0.16807
for x = 1
5C1*0.3¹(1-0.3)^5-1
= 5*0.3(0.7)⁴
= 5x0.3x0.2401
= 0.36015
for x = 2
5C2 * 0.3² * (1-0.3) ^5-2
= 0.30870
for x = 3
5C3 * 0.3³ * (1-0.3) ^ 5-3
= 10 * 0.027 * 0.7²
= 0.1323
for x = 4
5C4 * 0.3⁴ (1-0.3) ^ 5-4
= 5 * 0.0081 * 0.7
= 0.02835
for x = 5
5C5 *0.3⁵ (1- 0.3) ^5-5
= 1*0.00243*0,7⁰
= 0.00243
c. E[X] = N*p = 5*0.3 = 1.5
var[X] = np(1-p) = 5*0.3*0.7 = 1.05
d. 20/9 = 2.222
so we have that if x is greater than or equal to 3, cost will exceed 20
p(x=3) + p(x=4) + p(x=5)
= 0.1323 + 0.02835 + 0.00243
probability = 0.16308
E[C] = 1.5 * 9 = 13.5
VAR[C] = 1.05 * 9 = 9.45