true.we can find measurements of angle using cosec,sec and cot
Answer:
a) ![E(Y) =\sum_{i=1}^n y_i P(Y=y_i)](https://tex.z-dn.net/?f=%20E%28Y%29%20%3D%5Csum_%7Bi%3D1%7D%5En%20y_i%20P%28Y%3Dy_i%29)
![E(Y) = 0*0.6 +1*0.2 +2*0.15 +3*0.15= 0.95](https://tex.z-dn.net/?f=%20E%28Y%29%20%3D%200%2A0.6%20%2B1%2A0.2%20%2B2%2A0.15%20%2B3%2A0.15%3D%200.95)
b) ![E(80Y^2) = 80 E(Y^2)](https://tex.z-dn.net/?f=%20E%2880Y%5E2%29%20%3D%2080%20E%28Y%5E2%29)
![E(Y^2) =\sum_{i=1}^n y^2_i P(Y=y_i)](https://tex.z-dn.net/?f=%20E%28Y%5E2%29%20%3D%5Csum_%7Bi%3D1%7D%5En%20y%5E2_i%20P%28Y%3Dy_i%29)
![E(Y) = 0^2*0.6 +1^2*0.2 +2^2*0.15 +3^2*0.15= 2.15](https://tex.z-dn.net/?f=%20E%28Y%29%20%3D%200%5E2%2A0.6%20%2B1%5E2%2A0.2%20%2B2%5E2%2A0.15%20%2B3%5E2%2A0.15%3D%202.15)
![E(80Y^2) = 80 E(Y^2)= 80*2.15 =172](https://tex.z-dn.net/?f=%20E%2880Y%5E2%29%20%3D%2080%20E%28Y%5E2%29%3D%2080%2A2.15%20%3D172)
Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
For this case we have defined the following random variable Y="number of moving violations for which the individual was cited during the last 3 years. "
And we have the distribution for Y given:
y 0 1 2 3
P(y) 0.6 0.2 0.15 0.15
Part a
For this case the expected value is given by:
![E(Y) =\sum_{i=1}^n y_i P(Y=y_i)](https://tex.z-dn.net/?f=%20E%28Y%29%20%3D%5Csum_%7Bi%3D1%7D%5En%20y_i%20P%28Y%3Dy_i%29)
And if we replace the values given we have:
![E(Y) = 0*0.6 +1*0.2 +2*0.15 +3*0.15= 0.95](https://tex.z-dn.net/?f=%20E%28Y%29%20%3D%200%2A0.6%20%2B1%2A0.2%20%2B2%2A0.15%20%2B3%2A0.15%3D%200.95)
Part b
For this case we have defined a new random variable
representing a subcharge, and we want to find the expected amount for this random variable, using properties of expected value we have:
![E(80Y^2) = 80 E(Y^2)](https://tex.z-dn.net/?f=%20E%2880Y%5E2%29%20%3D%2080%20E%28Y%5E2%29)
And we can find
on this way:
![E(Y^2) =\sum_{i=1}^n y^2_i P(Y=y_i)](https://tex.z-dn.net/?f=%20E%28Y%5E2%29%20%3D%5Csum_%7Bi%3D1%7D%5En%20y%5E2_i%20P%28Y%3Dy_i%29)
And if we replace the values given we have:
![E(Y) = 0^2*0.6 +1^2*0.2 +2^2*0.15 +3^2*0.15= 2.15](https://tex.z-dn.net/?f=%20E%28Y%29%20%3D%200%5E2%2A0.6%20%2B1%5E2%2A0.2%20%2B2%5E2%2A0.15%20%2B3%5E2%2A0.15%3D%202.15)
And then replacing we got:
![E(80Y^2) = 80 E(Y^2)= 80*2.15 =172](https://tex.z-dn.net/?f=%20E%2880Y%5E2%29%20%3D%2080%20E%28Y%5E2%29%3D%2080%2A2.15%20%3D172)
Answer:
a. always
Step-by-step explanation:
In a simple linear regression, the slope measures the steepness of the linear relationship between the two variables, it ranges from (-∞ to +∞ ) while the correlation coefficient measures the tightness of the linear relationship between the two variables, it ranges from (-1 to +1 ).
So the correlation coefficient and the slope will always have the same sign though their values might differ.
So, the smallest candle has a radius r = 0.5 and a height of h = 3.
now, the other two candles in the set, are scaled of 2(twice as large) and of 3(thrice as large).
therefore, the candle scaled at 2 has a radius of 2*0.5 or 1, and a height of 2*3 or 6.
and the last candle in the set scaled t 3 has a radius of 3*0.5 or 1.5, and a height of 3*3, or 9
Answer:
232 sq. m
Step-by-step explanation: