B. Fractions have to have the same denominator so
![\frac{9}{10} - \frac{2}{5} = \frac{9}{10} - \frac{4}{10}](https://tex.z-dn.net/?f=%20%5Cfrac%7B9%7D%7B10%7D%20%20-%20%20%5Cfrac%7B2%7D%7B5%7D%20%3D%20%20%5Cfrac%7B9%7D%7B10%7D%20-%20%20%5Cfrac%7B4%7D%7B10%7D%20)
which =
![\frac{5}{10}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B10%7D%20)
then you simplfy it to
![\frac{1}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20)
so it would be B
Answer:
a
![y(t) = y_o e^{\beta t}](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_o%20e%5E%7B%5Cbeta%20t%7D)
b
![x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%20x_o%20e%5E%7B%5Cfrac%7B-%5Calpha%20y_o%20%7D%7B%5Cbeta%20%7D%5Be%5E%7B-%5Cbeta%20t%7D%20-%201%5D%20%7D)
c
![\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }](https://tex.z-dn.net/?f=%5Clim_%7Bt%20%5Cto%20%5Cinfty%7D%20x%28t%29%20%3D%20x_oe%5E%7B%5Cfrac%7B-%5Calpha%20y_o%7D%7B%5Cbeta%20%7D%20%7D)
Step-by-step explanation:
From the question we are told that
![\frac{dy}{y} = -\beta dt](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7By%7D%20%3D%20%20-%5Cbeta%20dt)
Now integrating both sides
![ln y = \beta t + c](https://tex.z-dn.net/?f=ln%20y%20%20%3D%20%20%5Cbeta%20t%20%2B%20c)
Now taking the exponent of both sides
![y(t) = e^{\beta t + c}](https://tex.z-dn.net/?f=y%28t%29%20%3D%20%20e%5E%7B%5Cbeta%20t%20%2B%20c%7D)
=> ![y(t) = e^{\beta t} e^c](https://tex.z-dn.net/?f=y%28t%29%20%3D%20%20e%5E%7B%5Cbeta%20t%7D%20e%5Ec)
Let ![e^c = C](https://tex.z-dn.net/?f=e%5Ec%20%3D%20%20C)
So
![y(t) = C e^{\beta t}](https://tex.z-dn.net/?f=y%28t%29%20%3D%20C%20e%5E%7B%5Cbeta%20t%7D)
Now from the question we are told that
![y(0) = y_o](https://tex.z-dn.net/?f=y%280%29%20%3D%20%20y_o)
Hence
![y(0) = y_o = Ce^{\beta * 0}](https://tex.z-dn.net/?f=y%280%29%20%3D%20y_o%20%20%3D%20Ce%5E%7B%5Cbeta%20%2A%200%7D)
=> ![y_o = C](https://tex.z-dn.net/?f=y_o%20%3D%20C)
So
![y(t) = y_o e^{\beta t}](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_o%20e%5E%7B%5Cbeta%20t%7D)
From the question we are told that
![\frac{dx}{dt} = -\alpha xy](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%20%20%3D%20-%5Calpha%20xy)
substituting for y
![\frac{dx}{dt} = - \alpha x(y_o e^{-\beta t })](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%20%20%3D%20-%20%5Calpha%20x%28y_o%20e%5E%7B-%5Cbeta%20t%20%7D%29)
=> ![\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bx%7D%20%20%3D%20-%5Calpha%20y_oe%5E%7B-%5Cbeta%20t%7D%20dt)
Now integrating both sides
![lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c](https://tex.z-dn.net/?f=lnx%20%3D%20%5Calpha%20%5Cfrac%7By_o%7D%7B%5Cbeta%20%7D%20e%5E%7B-%5Cbeta%20t%7D%20%2B%20c)
Now taking the exponent of both sides
![x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}](https://tex.z-dn.net/?f=x%28t%29%20%3D%20e%5E%7B%5Calpha%20%5Cfrac%7By_o%7D%7B%5Cbeta%20%7D%20e%5E%7B-%5Cbeta%20t%7D%20%2B%20c%7D)
=> ![x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c](https://tex.z-dn.net/?f=x%28t%29%20%3D%20e%5E%7B%5Calpha%20%5Cfrac%7By_o%7D%7B%5Cbeta%20%7D%20e%5E%7B-%5Cbeta%20t%7D%20%7D%20e%5Ec)
Let ![e^c = A](https://tex.z-dn.net/?f=e%5Ec%20%20%3D%20%20A)
=> ![x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }](https://tex.z-dn.net/?f=x%28t%29%20%3DK%20e%5E%7B%5Calpha%20%5Cfrac%7By_o%7D%7B%5Cbeta%20%7D%20e%5E%7B-%5Cbeta%20t%7D%20%7D)
Now from the question we are told that
![x(0) = x_o](https://tex.z-dn.net/?f=x%280%29%20%3D%20%20x_o)
So
![x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }](https://tex.z-dn.net/?f=x%280%29%3Dx_o%20%3DK%20e%5E%7B%5Calpha%20%5Cfrac%7By_o%7D%7B%5Cbeta%20%7D%20e%5E%7B-%5Cbeta%20%2A%200%7D%20%7D)
=> ![x_o = K e^{\frac {\alpha y_o }{\beta } }](https://tex.z-dn.net/?f=x_o%20%3D%20K%20e%5E%7B%5Cfrac%20%7B%5Calpha%20y_o%20%20%7D%7B%5Cbeta%20%7D%20%7D)
divide both side by ![(K * x_o)](https://tex.z-dn.net/?f=%20%28K%20%2A%20x_o%29)
=> ![K = x_o e^{\frac {\alpha y_o }{\beta } }](https://tex.z-dn.net/?f=K%20%3D%20x_o%20e%5E%7B%5Cfrac%20%7B%5Calpha%20y_o%20%20%7D%7B%5Cbeta%20%7D%20%7D)
So
![x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }](https://tex.z-dn.net/?f=x%28t%29%20%3Dx_o%20e%5E%7B%5Cfrac%20%7B-%5Calpha%20y_o%20%20%7D%7B%5Cbeta%20%7D%20%7D%20%2A%20%20e%5E%7B%5Calpha%20%5Cfrac%7By_o%7D%7B%5Cbeta%20%7D%20e%5E%7B-%5Cbeta%20t%7D%20%7D)
=> ![x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }](https://tex.z-dn.net/?f=x%28t%29%3D%20x_o%20e%5E%7B%5Cfrac%7B-%5Calpha%20%2A%20y_o%20%7D%7B%5Cbeta%7D%20%2B%20%5Cfrac%7B%5Calpha%20y_o%7D%7B%5Cbeta%20%7D%20e%5E%7B-%5Cbeta%20t%7D%20%7D)
=> ![x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%20x_o%20e%5E%7B%5Cfrac%7B%5Calpha%20y_o%20%7D%7B%5Cbeta%20%7D%5Be%5E%7B-%5Cbeta%20t%7D%20-%201%5D%20%7D)
Generally as t tends to infinity ,
tends to zero
so
![\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }](https://tex.z-dn.net/?f=%5Clim_%7Bt%20%5Cto%20%5Cinfty%7D%20x%28t%29%20%3D%20x_oe%5E%7B%5Cfrac%7B-%5Calpha%20y_o%7D%7B%5Cbeta%20%7D%20%7D)
106=x+x+5+x+8
106=3x+13
3x=93
x=31
(the sides are 31, 36, 39 respectively)
Answer:
The probability that all are male of choosing '3' students
P(E) = 0.067 = 6.71%
Step-by-step explanation:
Let 'M' be the event of selecting males n(M) = 12
Number of ways of choosing 3 students From all males and females
![n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276](https://tex.z-dn.net/?f=n%28M%29%20%3D%2028C_%7B3%7D%20%3D%20%5Cfrac%7B28%21%7D%7B%2828-3%29%213%21%7D%20%3D%5Cfrac%7B28%20X%2027%20X%2026%7D%7B3%20X%202%20X%201%20%7D%20%3D%203%2C276)
Number of ways of choosing 3 students From all males
![n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220](https://tex.z-dn.net/?f=n%28M%29%20%3D%2012C_%7B3%7D%20%3D%20%5Cfrac%7B12%21%7D%7B%2812-3%29%213%21%7D%20%3D%5Cfrac%7B12%20X%2011%20X%2010%7D%7B3%20X%202%20X%201%20%7D%20%3D220)
The probability that all are male of choosing '3' students
![P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }](https://tex.z-dn.net/?f=P%28E%29%20%3D%20%5Cfrac%7Bn%28M%29%7D%7Bn%28S%29%7D%20%3D%20%5Cfrac%7B12%20C_%7B3%7D%20%7D%7B28%20C_%7B3%7D%20%7D)
![P(E) = \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}](https://tex.z-dn.net/?f=P%28E%29%20%3D%20%20%5Cfrac%7B12%20C_%7B3%7D%20%7D%7B28%20C_%7B3%7D%20%7D%20%3D%20%5Cfrac%7B220%7D%7B3276%7D)
P(E) = 0.067 = 6.71%
<u><em>Final answer</em></u>:-
The probability that all are male of choosing '3' students
P(E) = 0.067 = 6.71%
If we round it
1) 7 people
2) 4 people
3) 5 people