Answer:
g= -2
Step-by-step explanation:
<span>(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer
than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
</span>
Answer:
4 cm
which is none of the listed answers
Step-by-step explanation:
∛64 = 4
proof:
4x4x4 = 64
For the first digit you are choosing from 2 digits
for the second digit you are choosing from 2 digits
for the third digit you are choosing from 2 digits
2*2*2=8
111
110
011
101
000
001
100
010
However......
A number doesn't usually start with a 0 or 0s.
Therefore, if you want 3-digit numbers and not just permutations using 0 and 1, then you must eliminate
011, 000, 001, and 010
Seeing that the first digit can't be 0, you choose from 1, then 2, and then 2 digits again; 1*2*2=4 numbers
You choose which answer best suits your problem.
Answer:
-12
Step-by-step explanation:
