He doesn’t have enough for both to be on at a time but if the cars are on without the train the cars will take up 2.4 meters leaving them with an extra .1 meters of length
Answer:
Option 4
Step-by-step explanation:
4+13= 17
4/13=0.307
4x13=52
4-13=-9
Answer:
293.38 pounds
Step-by-step explanation:
We are given that
Distance between poles=35 feet
Weight of cable=10.4 per linear foot
We have to find the weight of the cable.
Differentiate w.r.t
Let 
![s=\frac{2}{0.0225}\times\frac{2}{3}[t^{\frac{3}{2}}]^{17.5}_{0}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B2%7D%7B0.0225%7D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%5Bt%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5D%5E%7B17.5%7D_%7B0%7D)
![s=2\times \frac{2}{3\times0.0225}[(1+0.0255x)^{\frac{3}{2}]^{17.5}_{0}](https://tex.z-dn.net/?f=s%3D2%5Ctimes%20%5Cfrac%7B2%7D%7B3%5Ctimes0.0225%7D%5B%281%2B0.0255x%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%5D%5E%7B17.5%7D_%7B0%7D)
s=28.21
Weight of cable=
pound
Answer:
Our answer is 0.8172
Step-by-step explanation:
P(doubles on a single roll of pair of dice) =(6/36) =1/6
therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)
=1-(1-1/6)3 =91/216
for 12 players this follows binomial distribution with parameter n=12 and p=91/216
probability that at least 4 of the players will get “doubles” at least once =P(X>=4)
=1-(P(X<=3)
=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)
=1-0.1828
=0.8172
