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3 years ago

Given that (ax^2 + bx + 3) (x + d) = x^3 + 6x^2 + 11x + 12 a + 2b - d = ?

Mathematics

1 answer:

Mariulka [41]3 years ago

8 0

Answer:

Let's solve for a.

(ax2+bx+3)(x+d)=x3+6x2+11x+12a+2b−d

Step 1: Add -12a to both sides.

adx2+ax3+bdx+bx2+3d+3x+−12a=x3+6x2+12a+2b−d+11x+−12a

adx2+ax3+bdx+bx2−12a+3d+3x=x3+6x2+2b−d+11x

Step 2: Add -bdx to both sides.

adx2+ax3+bdx+bx2−12a+3d+3x+−bdx=x3+6x2+2b−d+11x+−bdx

adx2+ax3+bx2−12a+3d+3x=−bdx+x3+6x2+2b−d+11x

Step 3: Add -bx^2 to both sides.

adx2+ax3+bx2−12a+3d+3x+−bx2=−bdx+x3+6x2+2b−d+11x+−bx2

adx2+ax3−12a+3d+3x=−bdx−bx2+x3+6x2+2b−d+11x

Step 4: Add -3d to both sides.

adx2+ax3−12a+3d+3x+−3d=−bdx−bx2+x3+6x2+2b−d+11x+−3d

adx2+ax3−12a+3x=−bdx−bx2+x3+6x2+2b−4d+11x

Step 5: Add -3x to both sides.

adx2+ax3−12a+3x+−3x=−bdx−bx2+x3+6x2+2b−4d+11x+−3x

adx2+ax3−12a=−bdx−bx2+x3+6x2+2b−4d+8x

Step 6: Factor out variable a.

a(dx2+x3−12)=−bdx−bx2+x3+6x2+2b−4d+8x

Step 7: Divide both sides by dx^2+x^3-12.

a(dx2+x3−12)

dx2+x3−12

−bdx−bx2+x3+6x2+2b−4d+8x

dx2+x3−12

−bdx−bx2+x3+6x2+2b−4d+8x

dx2+x3−12

Answer:

−bdx−bx2+x3+6x2+2b−4d+8x/

dx2+x3−12

Step-by-step explanation:

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(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable $X\sim Poisson(\lambda = 0.62)$ .

The probability function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...$

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Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

= 1 - P (X = 0) - P (X = 1)

$=1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252$

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: $\lambda=7\times0.62=4.34$

Compute the probability of no deaths in a corps over 7 years as follows:

$P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130$

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

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Step-by-step explanation:

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