Answer:
4 days
Step-by-step explanation:
We need to use the probability functions of each of the intervals to know the Probability number and then use it in the expected value.
P(x>3 cargo ships) = 1-P(x<=3)
P(x>3) = 1-P(x<=3)
P(x>3) = 1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)]
![P(x>3) = 1 - [\frac{e^{-2}2^0}{01}+\frac{e^{-2}2^1}{11}+\frac{e^{-2}2^2}{21}+\frac{e^{-3}2^3}{31}]](https://tex.z-dn.net/?f=P%28x%3E3%29%20%3D%201%20-%20%5B%5Cfrac%7Be%5E%7B-2%7D2%5E0%7D%7B01%7D%2B%5Cfrac%7Be%5E%7B-2%7D2%5E1%7D%7B11%7D%2B%5Cfrac%7Be%5E%7B-2%7D2%5E2%7D%7B21%7D%2B%5Cfrac%7Be%5E%7B-3%7D2%5E3%7D%7B31%7D%5D)
P(x>3) = 1- [0.1353(1+2+2+1.33)]
P(x>3) = 1-0.856
P(x>3) = 0.1431
Als n=30, expected number is
E(x) =30*P(x>3)
E(x) = 30*0.1431
E(x) = 4.293
I expect 4.293 or 4 days per month the block being unable to hanlde all arriving ships
Okay so, I am assuming the model is smaller so it would be 0.34m long.
The model has to be 200 times smaller than the actual plane
Divide 68/200 and you get 0.34
Then to check your answer, multiply 0.34 by 200 which is 68 meaning
0.34:68 is equal to 1:200
Assuming the model is bigger (just in case) it would be 13,600m long.
The in this scenario, the model has to be 200 times bigger so it would be 68(200) which is 13,600
Hope this helps
Answer:
5 children
Step-by-step explanation:
$7.31 x2 = $14.62
$46- 14.62 =$32
5.66 divided by 5 = 28.3
5.66 divided by 6 = 33
and $33 is bigger than $31
So the greatest number of children tickets can be purchased is only 5.
1/2
3/8 - 7/8
Add 8 to the three and do 9-1 to get 11/8 and 8.
8-8=0
11/8 - 7/8 = 4/8
Simplify to 1/2
