Answer:
Determine if the following system can be approximated as a general form second order system, and if so, what is the approximate system. Include a plot of a step response comparison of the original system and its approximation. G(s) = s + 22/(s + 11) (s^2 + 2s + 4)
Step-by-step explanation:
Determine if the following system can be approximated as a general form second order system, and if so, what is the approximate system. Include a plot of a step response comparison of the original system and its approximation. G(s) = s + 22/(s + 11) (s^2 + 2s + 4)
Answer:
what are the dimensions of the track?
Step-by-step explanation:
Ln x + xe^y = 1
1/x + xf'(x)e^y + e^y = 0
f'(x) = (-e^y - 1/x)/xe^y
f'(1, 0) = (-e^0 - 1/1)/e^0 = (-1 - 1)/1 = -2
Let the required equation of tangent be y = mx + c, where y = 0, m = -2, x = 1
0 = -2(1) + c = -2 + c
c = 2
Therefore, required equation is y = -2x + 2
Answer:
The probability that the mean of this sample is less than 16.1 ounces of beverage is 0.0537.
Step-by-step explanation:
We are given that the average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.4 ounces.
A random sample of sixty-five 16-ounce beverage cans are selected
Let = <u><em>sample mean amount of a beverage</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean amount of a beverage = 16.18 ounces
= standard deviation = 0.4 ounces
n = sample of 16-ounce beverage cans = 65
Now, the probability that the mean of this sample is less than 16.1 ounces of beverage is given by = P( < 16.1 ounces)
P( < 16.1 ounces) = P( < ) = P(Z < -1.61) = 1 - P(Z 1.61)
= 1 - 0.9463 = <u>0.0537</u>
The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9591.