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3 years ago

Simplify 15x^2-18/6 A. 15x+3 B. X+3 C. 5x^2-18/2 D. 5x^2-6/2

Mathematics

1 answer:

ziro4ka [17]3 years ago

3 0

Answer:

D. 5x^2-6/2

Step-by-step explanation:

5x^2-18/6 , 18/6 is 6/2

5x^2-6/2 ,

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In a city school, 60% of students have blue eyes, 55% have dark hair, and 35% have blue eyes and dark hair. What is the probabil

gulaghasi [49]

Answer:

58%

Step-by-step explanation:

This is a problem of conditional probability.

Let A represent the event that student has dark hair.

So P(A) = 55% = 0.55

Let B represents the event that student has blue eyes.

So, P(B) = 60% = 0.60

Probability that student has blue eyes and dark hairs = P(A and B) = 35% = 0.35

We are to find the probability that a randomly selected student will have dark hair, given that the student has blue eyes. Using the given formula and values, we get:

$P(A|B)=\frac{P(A \cap B)}{P(B)}\\\\ P(A|B)=\frac{0.35}{0.60}\\\\ P(A|B)=0.58$

Therefore, there is 0.58 or 58% probability that the student will have dark hairs, given that the student has blue eyes.

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3 years ago

Solve the following differential equation: y" + y' = 8x^2

PtichkaEL [24]

Answer:

$y=y_p+y_h = \frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$

Step-by-step explanation:

Let's find a particular solution. We need a function of the form $y= ax^3+bx^2+cx+d$ such that

$y'= 3ax^2+2bx+c$ and

$y''= 6ax+2b$

$y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2$

then, 3a= 8, 2b+6a =0 and c+2b = 0. With the first equation we obtain

a = 8/3 and replacing in the second equation 2b+6(8/3) = 2b + 16 = 0. Then, b = -8. Finally, c = -2(-8) = 16.

So, our particular solution is $y_p= \frac{8}{3}x^3-8x^2+16x$ .

Now, let's find the solution $y_p$ of the homogeneus equation $y''+y'=0$ with the method of constants coefficients. Let $y=e^{\lambda x}$

$y'=\lambda e^{\lambda x}$

$y''=\lambda^2e^{\lambda x}$

then $\lambda e^{\lambda x}+\lambda^2 e^{\lambda x} = 0$

$e^{\lambda x}(\lambda +\lambda^2)= 0$

$(\lambda +\lambda^2)= 0$

$\lambda (1+\lambda)= 0$

$\lambda =0$ and $\lambda)= -1$ .

So, $y_h = C_1 + C_2e^{-x}$ and the solution is

$y=y_p+y_h =\frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$ .

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