Answer:
Step-by-step explanation:
1)The probability of success in each of the 58 identical engine tests is p=0.92
n = 58
mean, u = np = 58×0.92 = 53.36
2) The only value that would be considered usual for this distribution is 91. This is because it is the only value between the minimum and maximum value
3) n = 546
p = 17/100 = 0.17
Mean = np = 546×0.17= 92.82
4) n = 1035
p = 36/100 = 0.36
np = 1035 × 0.36 = 372.6
5) The probability of success is 0.2.
p = 0.2
q= 1-p = 1-0.2 = 0.8
n = 35
standard deviation =
√npq = √35×0.2×0.8 = 2.34
6) p = 0.25
q = 1-0.25 = 0.75
n = 5
Variance = npq = 5×0.25×0.75 = 0.9
7) n = 982
p = 0.431
q = 1 - p = 1 - 0.431 = 0.569
Variance = npq = 982×0.431×0.569= 240.8
8) n = 500
p = 84/100 = 0.84
q = 1-0.84 = 0.16
Standard deviation = √npq
Standard deviation = √500×0.84×0.16 = 8.2
The fifth term of the geometric sequence would be 0.972
Commutative property of addition is when 3+9=12 and 9+3=12.
<span>suppose the speed = x mi/h
Overspeed cost = 115 + x - 65 = 50 + x ....... 1
Overspeed Cost = 50 + (x -65)*10 = 10x - 600 ..........2
from [1] and [2], we get,
10x - 600 = 50 + x
10x -x = 50 +600
9x = 650
x = 72.22 mi/h
hope it helps
</span>
Answer:
<h2>p(B) =
8310</h2>
Step-by-step explanation:
We will use the addition rule of probability of two events to solve the question. According to the rule given two events A and B;
p(A∪B) = p(A)+p(B) - p(A∩B) where;
A∪B is the union of the two sets A and B
A∩B is the intersection between two sets A and B
Given parameters
P(A)=15
P(A∪B)=1225
P(A∩B)=7100
Required
Probability of event B i.e P(B)
Using the expression above to calculate p(B), we will have;
p(A∪B) = p(A)+p(B) - p(A∩B)
1225 = 15+p(B)-7100
p(B) = 1225-15+7100
p(B) = 8310
Hence the missing probability p(B) is 8310.
