9514 1404 393
Answer:
a) E = 6500 -50d
b) 5000 kWh
c) the excess will last only 130 days, not enough for 5 months
Step-by-step explanation:
<u>Given</u>:
starting excess (E): 6500 kWh
usage: 50 kWh/day (d)
<u>Find</u>:
a) E(d)
b) E(30)
c) E(150)
<u>Solution</u>:
a) The exces is linearly decreasing with the number of days, so we have ...
E(d) = 6500 -50d
__
b) After 30 days, the excess remaining is ...
E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days
__
c) After 150 days, the excess remaining would be ...
E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system
The supply is not enough to last for 5 months.
Answer:
0.048 is the probability that more than 950 message arrive in one minute.
Step-by-step explanation:
We are given the following information in the question:
The number of messages arriving at a multiplexer is a Poisson random variable with mean 15 messages/second.
Let X be the number of messages arriving at a multiplexer.
Mean = 15
For poison distribution,
Mean = Variance = 15

From central limit theorem, we have:
where n is the sample size.
Here, n = 1 minute = 60 seconds
P(x > 950)
Calculation the value from standard normal z table, we have,

0.048 is the probability that more than 950 message arrive in one minute.
It won't. Both are increasing by 1.5 every year. If you graph that, the lines will be parallel. Parallel lines have no intercept.
Answer: 1649
Step-by-step explanation: