Answer:
D
Step-by-step explanation:
Answer:
The answers are a) 1/3 and b) 1/3
Step-by-step explanation:
we will consider t to be the arrival time variable in minutes. We will have it run from 0 min to 30 min omitting the 7 hours , which won't change the results since the PDF we are about to calculate is a constant.
So the PDF of the arrival time is a constant and the since the area under this PDF distribution should be equal to 1 so, Let the height of the constant distribution equal to c, so c*30 (which is equal to the total probability) would be the area under the distribution, but this area should be equal to 1, So that gives us c=1/30 which is the value of the constant PDF for all corresponding arrival times.
part a) of the question asks for the probability that the passengers wait less than 5 minute. The passengers would have to wait for the bus 5 min or late if they arrive between the times (10 - 15 )min and (25 - 30) min. So we will have to integrate the PDF corresponding to these times and then we will will just have to add the probabilities calculated as give below,

Now part b) of the question asks for the probability that the passenger waits for more than 10 minutes. Which can be calculated by noting that that can only happen if the passenger arrives between the times (0 - 5) min and (15 - 20) min. So we will have to integrate the PDF corresponding to these times and then we will will just have to add the probabilities calculated as give below,

The answer to part b) is 1/3.
Answer:
d = 65°
w = 55°
Step-by-step explanation:
m∠DAC +∠m∠CAB = 60
m∠DAC + 35 = 60
m∠DAC = 25
m∠DAC + d = 90
25 + d = 90
d = 65°
m∠GAF + w + d = 180
60 + w + 65 = 180
w + 125 = 180
w = 55°
Answer:
1.649 approximately 2
Step-by-step explanation:
S.d = standard deviation = 0.5
Time taken = lead time = 2 weeks
Mean = demand for week = 5 boxes
We are required to find the safety stock to maintain at 99% service level.
At 99% level, the Z value is equal to 2.326.
Therefore,
Safety stock = z × s.d × √Lt
= 2.326 × 0.5 x √2
= 1.649
Which is approximately 2.
The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

In this case, we have
<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> ) ==> d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )
<em>y(t)</em> = 5 - 2<em>t</em> ==> d<em>y</em>/d<em>t</em> = -2
and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then




