Answer:
The distribution is 
Solution:
As per the question:
Total no. of riders = n
Now, suppose the 
is the time between the departure of the rider i - 1 and i from the cable car.
where
= independent exponential random variable whose rate is 
The general form is given by:
(a) Now, the time distribution of the last rider is given as the sum total of the time of each rider:
Now, the sum of the exponential random variable with with rate is given by:
<span>5.823 x 103 i hope it right ???</span>
The answer will be six x plus four
Answer:
tan(2u)=[4sqrt(21)]/[17]
Step-by-step explanation:
Let u=arcsin(0.4)
tan(2u)=sin(2u)/cos(2u)
tan(2u)=[2sin(u)cos(u)]/[cos^2(u)-sin^2(u)]
If u=arcsin(0.4), then sin(u)=0.4
By the Pythagorean Identity, cos^2(u)+sin^2(u)=1, we have cos^2(u)=1-sin^2(u)=1-(0.4)^2=1-0.16=0.84.
This also implies cos(u)=sqrt(0.84) since cosine is positive.
Plug in values:
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.84-0.16]
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.68]
tan(2u)=[(0.4)(sqrt(0.84)]/[0.34]
tan(2u)=[(40)(sqrt(0.84)]/[34]
tan(2u)=[(20)(sqrt(0.84)]/[17]
Note:
0.84=0.04(21)
So the principal square root of 0.04 is 0.2
Sqrt(0.84)=0.2sqrt(21).
tan(2u)=[(20)(0.2)(sqrt(21)]/[17]
tan(2u)=[(20)(2)sqrt(21)]/[170]
tan(2u)=[(2)(2)sqrt(21)]/[17]
tan(2u)=[4sqrt(21)]/[17]
Answer:
6 ¼ hours of Daylight
Step-by-step explanation:
- a day = 24 hours
- you know that there was 17 ¾ hours of no day light
- 17 ¾ = 17.75
- 24 - 17.75 = Daylight hours
- daylight hours = 6.25
