We will assume that the trains pass by his house following a uniform distribution with values between 0 and 24. The probability of a train passing on a 9-hour time period is 9/24 = 3/8 = 0.375. Lets call Y the amount of trains passing by his house during that 9-hour period. Y follows a Binomail distribution with parameters 22 and 0.375.

P(Y ≤ 5) = P(Y = 0) + P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) =

${22 \choose 0} * 0.375^0*(1-0.375)^{22} + {22 \choose 1}*0.375^1*(1-0.375)^{21} +\\\\{22 \choose 2} * 0.375^2*(1-0.375)^{20} + {22 \choose 3}*0.375^3*(1-0.375)^{19} + \\{22 \choose 4} * 0.375^4*(1-0.375)^{18} + {22 \choose 5}*0.375^5*(1-0.375)^{17} = 0.11069$

I hope that works for you!

4 0

2 years ago

he triangles below are similar. Triangle Y X Z. Side Y X is 6 centimeters, X Z is 7 centimeters, and Z Y is 3 centimeters. Trian

MissTica

YX=6
XZ=7
ZY=3

And

AB=12
BC=6
AC=14

Check

AB/YX=BC/ZY=AC/XZ
12/6=6/3=14/7
2=2=2

Hence

ABC=XYZ

ZYX≈CBA

Option C

6 0

1 year ago