Jermaine is saving $60 per week for a new guitar. The amount he has saved, s, is represented by
1 answer:
You might be interested in
The amplitude of the sine wave with RMS value of 220 V is
A = 220√2 volts.
The sine waveform is
v(t) = 220√2 sin(2πft)
where
f = 50 Hz, the frequency.
The period is
T = 1/f = 1/50 = 0.02 s
Use a graphical solution (shown below) to determine the number of times that v(t) = 220 in the interval t = [0, 0.02] s.
There are 2 instances when the voltage is 220 V in the interval t =[0, 0.02] s.
Note that 1 second is an integral multiple of 0.02 seconds.
Therefore in the interval [0,1], the number of instances when v(t) = 220 V is
(1/0.02)*2 = 100
Answer: 100
A = 220√2 volts.
The sine waveform is
v(t) = 220√2 sin(2πft)
where
f = 50 Hz, the frequency.
The period is
T = 1/f = 1/50 = 0.02 s
Use a graphical solution (shown below) to determine the number of times that v(t) = 220 in the interval t = [0, 0.02] s.
There are 2 instances when the voltage is 220 V in the interval t =[0, 0.02] s.
Note that 1 second is an integral multiple of 0.02 seconds.
Therefore in the interval [0,1], the number of instances when v(t) = 220 V is
(1/0.02)*2 = 100
Answer: 100
Answer:
B
Step-by-step explanation:
The best - well, only way- is to check a few points.
Namely -1 (unless your eyesight is really poor!), 0, 1, 2, and 3.
Now you can mark all these points in each graph (well, you could if they were on paper and not on a screen) and see which one of the lines passes through all of them. Spoiler alert, it's the B graph.
A represents , B is the one you want, C is
and D looks like
Answer:
Step-by-step explanation:
given that certain tubes manufactured by a company have a mean lifetime of 800 hours and a standard deviation of 60 hours.
Sample size n =16
Std error of sample mean =
x bar follows N(800, 15)
the probability that a random sample of 16 tubes taken from the group will have a mean lifetime
(a) between 790 and 810 hours,
=
(b) less than 785 hours
, (c) more than 820 hours,
(d) between 770 and 830 hours
=