There are three questions related to this problem.
First, the probability of the mail will arrive after 2:30
PM
<span>Find the z-score of 2:30 which is 30 minutes after 2:00.</span>
<span>
z(2:30) = (2:30 – 2:00)/15 = -30/15 = -2</span>
<span>
P(x < 2:30) = P(z<-2) = 0.0228</span>
<span>
</span>
Second, the probability of the mail will arrive at 1:36
PM
<span>Find the z-score of 1:36 which is 24 minutes before 2:00.</span>
<span>
z(1:36) = (1:36 – 2:00)/15 = -24/15 = -1.6</span>
<span>
P(x < 1:36) = P(z<-1.6) = 0.0548</span>
Lastly, the probability of the mail will arrive between 1:48
PM and 2:09 PM
Find the z-score of 1:46 and 2:09 PM which will result to
a z value of 0.034599
<span>P(1:48 < x < 2:09) = P(z<0.034599) = 0.5138</span>
Answer:
see explanation
Step-by-step explanation:
p(- 3, 6 ) → p'(- 4, 8 )
Note that the x- coordinate of p' is 1 less than p and the y- coordinate of p' is 2 more than p, thus the translation is
![\left[\begin{array}{ccc}-1\\2\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%5C%5C2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Answer:
31.747 GW
Step-by-step explanation:
The total cell capacity (C) manufactured in the time period is the integral of the rate of manufacture. So, we have ...

About 31.747 GW of generating capacity was manufactured in that time interval.
See tha attached pdf file and let me know how useful this answer is.
I = PRT.....rearrange = I / PR = T
I = 450
R = 7.5%...turn to decimal = 0.075
P = 2400
I / (PR) = T
450 / (2400 * 0.075) = T
450 / 180 = T
2.5 = T...so time is 2.5 years, or 2 1/2 years