Answer:
Therefore the auxiliary solution is ![y=A e^{5x}+Be^{-4x}](https://tex.z-dn.net/?f=y%3DA%20e%5E%7B5x%7D%2BBe%5E%7B-4x%7D)
Therefore
are linearly independent
Step-by-step explanation:
Given, the differential equation is
y"-y'-20 y=0
Let
be the solution of the above differential equation.
y'=
and ![y"= m^2e^{mx}](https://tex.z-dn.net/?f=y%22%3D%20m%5E2e%5E%7Bmx%7D)
Then the above differential equation becomes
![m^2e^{mx}-me^{mx}-20 e^{mx}=0](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmx%7D-me%5E%7Bmx%7D-20%20e%5E%7Bmx%7D%3D0)
![\Rightarrow e^{mx}(m^2-m-20)=0](https://tex.z-dn.net/?f=%5CRightarrow%20e%5E%7Bmx%7D%28m%5E2-m-20%29%3D0)
![\Rightarrow (m^2-m-20)=0](https://tex.z-dn.net/?f=%5CRightarrow%20%28m%5E2-m-20%29%3D0)
![\Rightarrow m^2-5m+4m-20=0](https://tex.z-dn.net/?f=%5CRightarrow%20m%5E2-5m%2B4m-20%3D0)
![\Rightarrow m(m-5) +4(m-5)=0](https://tex.z-dn.net/?f=%5CRightarrow%20m%28m-5%29%20%2B4%28m-5%29%3D0)
![\Rightarrow (m-5)(m+4)=0](https://tex.z-dn.net/?f=%5CRightarrow%20%28m-5%29%28m%2B4%29%3D0)
![\Rightarrow m=5,-4](https://tex.z-dn.net/?f=%5CRightarrow%20m%3D5%2C-4)
If two roots of m are real and distinct then the auxiliary solution is
[where a and b are two roots of m]
Therefore the auxiliary solution is ![y=A e^{5x}+Be^{-4x}](https://tex.z-dn.net/?f=y%3DA%20e%5E%7B5x%7D%2BBe%5E%7B-4x%7D)
Wronskian
![W(e^{-4x},e^{5x})=\left[\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right]](https://tex.z-dn.net/?f=W%28e%5E%7B-4x%7D%2Ce%5E%7B5x%7D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%5E%7B-4x%7D%26e%5E%7B5x%7D%5C%5C-4e%5E%7B-4x%7D%265e%5E%7B5x%7D%5Cend%7Barray%7D%5Cright%5D)
![=5e^{-4x}e^{5x}-e^{5x}(-4e^{-4x})](https://tex.z-dn.net/?f=%3D5e%5E%7B-4x%7De%5E%7B5x%7D-e%5E%7B5x%7D%28-4e%5E%7B-4x%7D%29)
≠0
Therefore
are linearly independent.[ ∵W≠0]
Answer: the probability it will come up heads 25 or fewer times is 0.019
Step-by-step explanation:
Given that;
n = 50
p = 0.65
so, q = 1 - p = 0.35
np = 50 × 0.65 = 32.5 ≥ 10
nq = 50 × 0.35 = 17.5 ≥ 10
so, we need to use Normal Approximation for the Binomial Distribution
μ = np = 50 × 0.65 = 32.5
σ = √(npq) = √( 50 × 0.65 × 0.35 ) = 3.3726
now, the probability that it will come up heads 25 or few times will be;
⇒ P( x≤25)
{using continuity correction}
⇒ P[ z < (25.5 - 32.5)/3.3726 ]
⇒ P[ z < -2.0755 ]
using z-table
= 0.01923 ≈ 0.019 { 3 decimal places}
Therefore the probability it will come up heads 25 or fewer times is 0.019
Answer:
7, 10 and 15
Step-by-step explanation:
Given
--- Students already enrolled
![Total = 10\ (less)](https://tex.z-dn.net/?f=Total%20%3D%2010%5C%20%28less%29)
Required
Determine the additional number of students needed for the teacher not to cancel the class
First, we need to calculate the number of students for the teacher to cancel the class.
Represent this with x.
So, we have:
![S_1 + x < Total](https://tex.z-dn.net/?f=S_1%20%2B%20x%20%3C%20Total)
Make x the subject
![x < Total - S_1](https://tex.z-dn.net/?f=x%20%3C%20Total%20-%20S_1)
Substitute values for Total and S1
![x < 10 - 3](https://tex.z-dn.net/?f=x%20%3C%2010%20-%203)
![x < 7](https://tex.z-dn.net/?f=x%20%3C%207)
This means that if less than 7 additional students come to the class, the teacher will cancel the class.
So, for the teacher not to cancel the class, the students must be 7 or greater (i.e. 7, 8, 9, 10....)
54.16 is more precise because it has more digits