57.10
70.9
these are only the answers for the first two problems
Answer:
Continuous random variables: c and e
Discrete random variables: a, b, d
Step-by-step explanation:
We have to identify whether the random variable is discrete or continuous.
- A discrete variable is a variable whose value is obtained by counting.
- A continuous random variable X takes all values in a given interval of numbers.
- Thus, a continuous variable can have values in decimals but a discrete random variable cannot take values in decimals.
a. The number of statistics students now reading a book.
Discrete random variable since number of students cannot take decimal values.
b. The number of textbook authors now sitting at a computer.
Discrete random variable since number of textbooks cannot be expressed in decimals but counted.
c. The exact time it takes to evaluate 27 plus 72.
It is a continuous random variable as it may take all values within an interval of time.
d. The number of free dash throw attempts before the first shot is made.
It is a discrete random variable since the number of throws can always be whole number.
e. The time it takes to fly from City Upper A to City Upper B.
Time is a continuous random variable.
Answer:
Please read the complete procedure below:
Step-by-step explanation:
You have the following initial value problem:
a) The algebraic equation obtain by using the Laplace transform is:
![L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\](https://tex.z-dn.net/?f=L%5By%27%5D%2B2L%5By%5D%3D4L%5Bt%5D%5C%5C%5C%5CL%5By%27%5D%3DsY%28s%29-y%280%29%5C%20%5C%20%5C%20%5C%20%281%29%5C%5C%5C%5CL%5Bt%5D%3D%5Cfrac%7B1%7D%7Bs%5E2%7D%5C%20%5C%20%5C%20%5C%20%5C%20%282%29%5C%5C%5C%5C)
next, you replace (1) and (2):
(this is the algebraic equation)
b)
![sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)}](https://tex.z-dn.net/?f=sY%28s%29%2B2Y%28s%29-8%3D%5Cfrac%7B4%7D%7Bs%5E2%7D%5C%5C%5C%5CY%28s%29%5Bs%2B2%5D%3D%5Cfrac%7B4%7D%7Bs%5E2%7D%2B8%5C%5C%5C%5CY%28s%29%3D%5Cfrac%7B4%2B8s%5E2%7D%7Bs%5E2%28s%2B2%29%7D)
(this is the solution for Y(s))
c)
![y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}](https://tex.z-dn.net/?f=y%28t%29%3DL%5E%7B-1%7DY%28s%29%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%2B%5Cfrac%7B8%7D%7Bs%2B2%7D%5D%5C%5C%5C%5C%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%5D%2BL%5E%7B-1%7D%5B%5Cfrac%7B8%7D%7Bs%2B2%7D%5D%5C%5C%5C%5C%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%5D%2B8e%5E%7B-2t%7D)
To find the inverse Laplace transform of the first term you use partial fractions:
Thus, you have:
![y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}](https://tex.z-dn.net/?f=y%28t%29%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%5D%2B8e%5E%7B-2t%7D%5C%5C%5C%5Cy%28t%29%3DL%5E%7B-1%7D%5B%5Cfrac%7B-1%7D%7Bs%7D%2B%5Cfrac%7B2%7D%7Bs%5E2%7D%5D%2BL%5E%7B-1%7D%5B%5Cfrac%7B1%7D%7Bs%2B2%7D%5D%2B8e%5E%7B-2t%7D%5C%5C%5C%5Cy%28t%29%3D-1%2B2t%2Be%5E%7B-2t%7D%2B8e%5E%7B-2t%7D%3D-1%2B2t%2B9e%5E%7B-2t%7D)
(this is the solution to the differential equation)
No don't pleeeeeeassssseeer
Answer:
If two things have the same shape and size then they are <em>congruent</em> to one another. If they are only the same shape but not size, then they are similar.
