Answer:
Step-by-step explanation:
Given that three airlines serve a small town in Ohio. Airline A has 52% of all scheduled flights, airline B has 27% and airline C has the remaining 21%.
Their on-time rates are 83%, 66%, and 36%, respectively.
Airlines A B C
Flights 52% 27% 21% 100%
On time rates 83% 66% 36%
Flight*online rates 43.1600% 17.8200% 7.5600% 68.5400%
Thus the above table shows the product of no of flights and on time rates.
Prob it was airline A given it left on time
= 
Voters<span>: 226 It was around 445 I think, </span>as<span> you had </span>to work out<span> the </span>total<span>, which was 3+7 so</span><span> 20 was the sine rule question where the angle was 20.55 </span>degrees<span> </span><span>3360 </span><span>More women voted then men.</span>
250 miles ÷ 4.5 hours = 56 mph (rounded from 55.55555)
Answer:
point
Step-by-step explanation:
the dot has no connectives or no paths
Answer:
a) 0.32
b) 0.68
c) office or den
Step-by-step explanation:
Locations Probabilities
Adult bedroom 0.03
Child bedroom 0.15
Other bedroom 0.14
Office or den 0.40
Other rooms 0.28
a)
P(PC in bedroom)= P(PC in adult bedroom)+ P(PC in child bedroom)+ P(PC in other bedroom)
P(PC in bedroom)= 0.03+0.15+ 0.14
P(PC in bedroom)= 0.32.
Thus, the probability that a PC is in a bedroom is 0.32.
b)
P(PC is not in bedroom)= P(PC in Office or den)+ P(PC in Other rooms)
P(PC is not in bedroom)= 0.40+0.28
P(PC is not in bedroom)= 0.68.
Thus, the probability that a PC is not in a bedroom is 0.68.
c)
When a household is selected at random from households with a PC we would expect to find a PC in a room which has a greater probability of having PC.
The greater probability of room having a PC is of Office or den room with probability 0.40. So, when a household is selected at random from households with a PC we would expect to find a PC in a Office or den room.
