Lovely.
The car is traveling at 95 km/h, and the train is at 75 km/h.
Remember the train is 1 km long.
a)
With respect to the train the car is not moving as much as 75 km/h.
The car with respect to the train is moving as much as 95 - 75 = 20 km/h
Recall the train is 1 km long.
Time to cover the 1km long train = Distance / speed = 1 / 20 = (1/20) hour.
= (1/20)*60 minutes = 3 minutes.
So the car would pass the train in 3 minutes.
b)
Now if they are traveling in opposite directions;
the car with respect to the train = 75 + 95 = 170 km/h
Time to cover the 1km long train = Distance / speed = 1 / 170 = (1/170) hour.
<span>= (1/170)*60 minutes = (6/17) minutes
</span>
= (6/17)*60 seconds = 360/17 seconds ≈ 21.176 seconds.
Let X be the number of female employee. Let n be the sample size, p be the probability that selected employee is female.
It is given that 45% employee are female it mean p=0.45
Sample size n=60
From given information X follows Binomial distribution with n=50 and p=0.45
For large value of n the Binomial distribution approximates to Normal distribution.
Let p be the proportion of female employee in the given sample.
Then distribution of proportion P is normal with parameters
mean =p and standard deviation =
Here we have p=0.45
So mean = p = 0.45 and
standard deviation =
standard deviation = 0.0642
Now probability that sample proportions of female lies between 0.40 and 0.55 is
P(0.40 < P < 0.45) =
= P(-0.7788 < Z < 1.5576)
= P(Z < 1.5576) - P(Z < -0.7788)
= P(Z < 1.56) - P(Z < -0.78)
= 0.9406 - 0.2177
= 0.7229
The probability that the sample proportion of females is between 0.40 and 0.55 is 0.7229
The answer is B / the second option
K(x,y)>K’(-x,y-7)
F(x) = 1/4x
subtitute x = -5
f(-5) = 1/4 · (-5) = -5/4 = -1 1/4