I = PRT
T = 6
P = 850
I = 1.9% = 0.019
I = (850)(0.019)(6) = 96.90 <==
Answer:
The data can be modeled using linear equatins, because there's the same amount of population for fish A (+20 per year) and fish B (+30 per year) every year.
Answer:
0.0143
Step-by-step explanation:
In this question, we are asked to use the binomial distribution to calculate the probability that 10 or fewer passengers from a sample of MIT data project sample were on American airline flights.
We proceed as follows;
The probability that a passenger was an American flight is 15.5%= 15.55/100 = 0.155
Let’s call this probability p
The probability that he/she isn’t on the flight, let’s call this q
q =1 - p= 0.845
Sample size, n = 155
P(X < A) = P(Z < (A - mean)/standard deviation)
Mean = np
= 125 x 0.155
= 19.375
Standard deviation = √npq
= √ (125 x 0.155x 0.845)
= 4.0462
P(10 or fewer passengers were on American Airline flights) = P(X \leq 10)
= P(Z < (10.5 - 19.375)/4.0462)
= P(Z < -2.19)
= 0.0143
Answer:
The height of this rectangle is 8/10
Step-by-step explanation:
Please see attachment .
(7/2,-7) ther I hope it help
