She may conclude that in her hometown, a majority of people experience greater happiness thanks to positive relationships. However, in order to apply this to a greater population, she will have to use a larger sample of people.
We have by the intermediate value theorem that if a continuous function takes values both above and below zero at 2 points, there is a zero of the function in-between. We have that polynomials are continues. Let's calculate f(-6) and f(-5). f(-6)=-36 while f(-5)=-1. Thus, we cannot conclude that there is a root between them.
F(-2)=8, f(-1)=-1, so there is a flip; a zero must exist between them.
F(1)=-1, f(2)=20, so again there is a change of signs.
f(-5)=-1, f(-4)=14 so there is a root still.
We have that the only choice that does not have a root between the integers is choice a.
For the given function f(t) = (2t + 1) using definition of Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].
As given in the question,
Given function is equal to :
f(t) = 2t + 1
Simplify the given function using definition of Laplace transform we have,
L(f(t))s = 
= ![\int\limits^\infty_0[2t +1] e^{-st} dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%5Cinfty_0%5B2t%20%2B1%5D%20e%5E%7B-st%7D%20dt)
= 
= 2 L(t) + L(1)
L(1) = 
= (-1/s) ( 0 -1 )
= 1/s , ( s > 0)
2L ( t ) = 
= ![2[t\int\limits^\infty_0 e^{-st} - \int\limits^\infty_0 ({(d/dt)(t) \int\limits^\infty_0e^{-st} \, dt )dt]](https://tex.z-dn.net/?f=2%5Bt%5Cint%5Climits%5E%5Cinfty_0%20e%5E%7B-st%7D%20-%20%5Cint%5Climits%5E%5Cinfty_0%20%28%7B%28d%2Fdt%29%28t%29%20%5Cint%5Climits%5E%5Cinfty_0e%5E%7B-st%7D%20%5C%2C%20dt%20%29dt%5D)
= 2/ s²
Now ,
L(f(t))s = 2 L(t) + L(1)
= 2/ s² + 1/s
Therefore, the solution of the given function using Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].
Learn more about Laplace transform here
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Answer:
50:20:50
This is because you add 5+2+5=12
120/12=10
5x10=50 2x10=20 5x10=50
23/6 as a mixed fraction is 3(5/6)
