Looking at the set, we are given 18 elements. 17 is prime; it has only two factors: 1 and 17, since 1•17=17. So, the question is really asking what is the probability the numbers 1 or 17 is chosen. As mentioned earlier, 17 is prime, so there are two possible choices: 1 and 17.
P (probability) = possible outcomes / total outcomes
It is important to note that these events are “or” events, meaning that the probability can only be determined by choosing a 1 or a 17; you can’t randomly chose a 1 and 17 at the same time. So, the formula is:
P(A or B) = P(A) + P(B)
All this is saying is that given two possible outcomes, the probability occurs independent of each event; they don’t occur at the same time.
P(1 or 17) = P(1)/18 + P(1)/18
P(1 or 17) = 2/18
Since 17 is prime, it’s two and only factors are 1 and 17. The probability of randomly choosing a 1 or 17 is 2/18, meaning that there are 2 elements in the set out of a possible 18 elements that can be randomly chosen.
2/18 simplifies to 1/9
So, your answer is 1/9
54 = 2 × 3 × 3 × 3 ( as a product of its prime factor ).
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)
Answer:
The question is open ended as i suppose since there are infinitely many possibilities. one possible form of the solution can be 
Step-by-step explanation:
The base in this case is 5 while the exponent is simply the power to which the base is raised in which case it would be 2. Both values are positive and the exponent is less than the base.
<span>Whenever we are given one function and must calculate a funciton of the funciton, such as g(f(x)) in this case, we simply substitute the second function, f(x) in this case, in the first function, g(x) in this case, wherever the first function has a variable. Therefore,
g(f(x)) = (3x + 2)^2 + 1
g(f(x)) = 9x^2 + 12x + 4 + 1
g(f(x)) = 9x^2 + 12x + 5</span>
