Ok. Could I have a picture of the question? :)
We have to see how many times 130 fits into 1494.23:
![1494.23/130=11.4940....](https://tex.z-dn.net/?f=1494.23%2F130%3D11.4940....)
So now we have to multiply this by the amount of time needed for 130 meters:
11.4940*1=11.4940.
So the answer is 11.4940 minutes.
I can’t see the question b
Answer:
0.30
Step-by-step explanation:
Probability of stopping at first signal = 0.36 ;
P(stop 1) = P(x) = 0.36
Probability of stopping at second signal = 0.54;
P(stop 2) = P(y) = 0.54
Probability of stopping at atleast one of the two signals:
P(x U y) = 0.6
Stopping at both signals :
P(xny) = p(x) + p(y) - p(xUy)
P(xny) = 0.36 + 0.54 - 0.6
P(xny) = 0.3
Stopping at x but not y
P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06
Stopping at y but not x
P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24
Probability of stopping at exactly 1 signal :
P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30