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3 years ago

Completing the Square

Mathematics

1 answer:

Softa [21]3 years ago

6 0

Answer:

C.7, -2

Step-by-step explanation:

A quadratic equation is a polynomial of degree two. A quadratic equation is in the form:

ax² + bx + c = 0; where a, b, c are constants.

Quadratic equation can be solved using different methods, such as completing the square method, quadratic formula.

Solving the equation 2x² - 10x - 20 = 8 using completing the square:

2x² - 10x - 20 = 8

2x² - 10x = 28

Divide through by 2, to make the coefficient of x² to be 1:

x² - 5x = 14

Add to both sides, the square of half of the coefficient of x [i.e. (-5/2)² = 6.25]

x² - 5x + 6.25 = 14 + 6.25

(x - 2.5)² = 20.25

taking square root of both sides:

x - 2.5 = √20.25

x - 2.5 = ±4.5

x = 2.5 ± 4.5

x = 2.5 + 4.5; or x = 2.5 - 4.5

x = 7 or x = -2

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sergey [27]

Answer:

100

Step-by-step explanation:

Ratio is a fraction, so ratio of boys to girls would be written as [let boys be "b" and girls be "g"]:

$\frac{b}{g}=\frac{2}{7}$

Also, there are 250 more girls than boys, so we can write:

g = 250 + b

Lets substitute this into "g" of the ratio we wrote and cross mulitply and solve for b:

$\frac{b}{g}=\frac{2}{7}\\\frac{b}{250+b}=\frac{2}{7}\\7b=2(250+b)\\7b=500+2b\\5b=500\\b=\frac{500}{5}=100$

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4 years ago

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nataly862011 [7]

Step-by-step explanation:

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If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calcul

Wittaler [7]

Answer:

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

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This means that for 1 week, $\mu = \frac{3}{20} = 0.15$

Calculate the probability that there will not be more than one failure during a particular week.

Probability of at most one failure, so:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

Then

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.15}*0.15^{0}}{(0)!} = 0.8607$

$P(X = 1) = \frac{e^{-0.15}*0.15^{1}}{(1)!} = 0.1291$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.8607 + 0.1291 = 0.9898$

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

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3 years ago

What 8s the answer please I need help

Sphinxa [80]

Tha answer
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Ilya [14]

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