The answer is 5.4 × 10^-4
To prove that jill is wrong we just need an example of this;
2*3*5*7*11*13 = 30030 (this is the smallest number with 6 different prime numbers)
5953*5981*5987 = 2.13x10^11 (which is obviously a much bigger number)
this is enough to prove that jill is wrong
<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:
8x - 7.
Step-by-step explanation:
f(x) = 2x - 5,
g(x) = 4x -1,
substituting g(x) as x in f(x),
f(g(x)) = 2(4x - 1) - 5
= 8x - 7
Set 4 is the correct answer because if you multiply 10 by 10 + 20 times 20, you will get 400 and when you multiply 30 by 30 you also get 400 which is equal to 400 there for you answer would be set 4.
Hope I helped if I did click the thx button ;)
