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3 years ago

9

What is the probability that a randomly selected child from this group will have red hair, given that he has straight hair? (Ent

er your answer as a fraction.)

1 answer:

Anit [1.1K]3 years ago

5 0

Answer:

Your question was incomplete, as the values are not given.

I guess these are the values you are talking about.

Hair Type Brown Blond Black Red Totals

Wavy 20 5 15 3 43

Straight 80 15 65 12 172

Totals 100 20 80 15 215

Now w have to find out the probability of a child with red, straight hair.

Total no of children with straight hair are = 172

The children having straight red hair = 12

The probability of selected child = 12/172.

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Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

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4 years ago

Jaewon is 5 feet 3 inches tall. How many inches tall is Jaewon? O A. 53 inches B60 inches OC. 63 inches oc O D. 96 inches

nadezda [96]

Answer:

First, we determine how many inches in the board. (5 x 12) + 3. This gives you 63 total inches. You want three equal pieces so you divide 63 by 3 and the answer is 21 inches for each individual piece.

Step-by-step explanation:

First, we determine how many inches in the board. (5 x 12) + 3. This gives you 63 total inches. You want three equal pieces so you divide 63 by 3 and the answer is 21 inches for each individual piece.

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3 years ago

Find thevmissing value. please

11Alexandr11 [23.1K]

Answers are
y=69°
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3 years ago

Explain the different ways you can name an angle

nydimaria [60]

Answer:

When you say that, I think directly to Triangles. SO i am going to state that.

There is acute: Less than 90 degrees

There is Obtuse: more than 90 degrees

There is Right: 90 degrees

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4 years ago

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Need Help!!!!!!!!!!!!!!!!!!

marin [14]

Answer:

12

Step-by-step explanation:

Just add three

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3 years ago

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