Hello,
Your answer would be:
D- 3
Pm me for work
Plz mark me brainliest
Answer:
![P(A\textrm{ and }B)=\frac{3}{14}](https://tex.z-dn.net/?f=P%28A%5Ctextrm%7B%20and%20%7DB%29%3D%5Cfrac%7B3%7D%7B14%7D)
Step-by-step explanation:
Given:
![P(A)=\frac{4}{7}](https://tex.z-dn.net/?f=P%28A%29%3D%5Cfrac%7B4%7D%7B7%7D)
![P(B|A)=\frac{3}{8}](https://tex.z-dn.net/?f=P%28B%7CA%29%3D%5Cfrac%7B3%7D%7B8%7D)
We know that, conditional probability of B given that A has occurred is given as:
. Expressing this in terms of
, we get
![P(A\cap B)=P(B|A)\times P(A)](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%3DP%28B%7CA%29%5Ctimes%20P%28A%29)
Plug in the known values and solve for
. This gives,
![P(A\cap B)=P(B|A)\times P(A)\\P(A\cap B)=\frac{3}{8}\times \frac{4}{7}\\P(A\cap B)=\frac{12}{56}=\frac{3}{14}](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%3DP%28B%7CA%29%5Ctimes%20P%28A%29%5C%5CP%28A%5Ccap%20B%29%3D%5Cfrac%7B3%7D%7B8%7D%5Ctimes%20%5Cfrac%7B4%7D%7B7%7D%5C%5CP%28A%5Ccap%20B%29%3D%5Cfrac%7B12%7D%7B56%7D%3D%5Cfrac%7B3%7D%7B14%7D)
Therefore, the probability of events A and B occurring is
.
Answer:
![\therefore y_2(x)=-\frac{e^{-6x}}{8}](https://tex.z-dn.net/?f=%5Ctherefore%20y_2%28x%29%3D-%5Cfrac%7Be%5E%7B-6x%7D%7D%7B8%7D)
The general solution is
![y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}](https://tex.z-dn.net/?f=y%3Dc_1e%5E%7B2x%7D-c_2.%5Cfrac%7Be%5E%7B-6x%7D%7D%7B8%7D)
Step-by-step explanation:
Given differential equation is
y''-4y'+4y=0
and ![y_1(x)=e^{2x}](https://tex.z-dn.net/?f=y_1%28x%29%3De%5E%7B2x%7D)
To find the
we are applying the following formula,
![y_2(x)=y_1(x)\int \frac{e^{-\int P(x) dx}}{y_1^2(x)} \ dx](https://tex.z-dn.net/?f=y_2%28x%29%3Dy_1%28x%29%5Cint%20%5Cfrac%7Be%5E%7B-%5Cint%20P%28x%29%20dx%7D%7D%7By_1%5E2%28x%29%7D%20%5C%20dx)
The general form of equation is
y''+P(x)y'+Q(x)y=0
Comparing the general form of the differential equation to the given differential equation,
So, P(x)= - 4
![\therefore y_2(x)=e^{2x}\int \frac{e^{-\int 4dx}}{(e^{2x})^2}dx](https://tex.z-dn.net/?f=%5Ctherefore%20y_2%28x%29%3De%5E%7B2x%7D%5Cint%20%5Cfrac%7Be%5E%7B-%5Cint%204dx%7D%7D%7B%28e%5E%7B2x%7D%29%5E2%7Ddx)
![=e^{2x}\int \frac{e^{-4x}}{e^{4x}}dx](https://tex.z-dn.net/?f=%3De%5E%7B2x%7D%5Cint%20%5Cfrac%7Be%5E%7B-4x%7D%7D%7Be%5E%7B4x%7D%7Ddx)
![=e^{2x}\int e^{-4x-4x} \ dx](https://tex.z-dn.net/?f=%3De%5E%7B2x%7D%5Cint%20e%5E%7B-4x-4x%7D%20%5C%20dx)
![=e^{2x}\int e^{-8x} \ dx](https://tex.z-dn.net/?f=%3De%5E%7B2x%7D%5Cint%20e%5E%7B-8x%7D%20%5C%20dx)
![=e^{2x}. \frac{e^{-8x}}{-8}](https://tex.z-dn.net/?f=%3De%5E%7B2x%7D.%20%5Cfrac%7Be%5E%7B-8x%7D%7D%7B-8%7D)
![=-\frac{e^{-6x}}{8}](https://tex.z-dn.net/?f=%3D-%5Cfrac%7Be%5E%7B-6x%7D%7D%7B8%7D)
![\therefore y_2(x)=-\frac{e^{-6x}}{8}](https://tex.z-dn.net/?f=%5Ctherefore%20y_2%28x%29%3D-%5Cfrac%7Be%5E%7B-6x%7D%7D%7B8%7D)
The general solution is
![y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}](https://tex.z-dn.net/?f=y%3Dc_1e%5E%7B2x%7D-c_2.%5Cfrac%7Be%5E%7B-6x%7D%7D%7B8%7D)
Answer:
Step-by-step explanation:
We don't have any statements from which to choose, but I did ind a couple of things that might be useful to you. Solving the system simultaneously for a, b, and c gives back a quadratic that is
and it has a vertex of (5, -.8). It opens upwards and is very wide open, which means that its vertex is a minimum. It has real zeros of x = 7 and x = 3, and a y-intercept of (0, 4.2). Not sure what else there is to tell you so hopefully I covered something of importance!
The correct answers would be the letter C