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2 years ago

*2.) Which trinomial is equivalent to 4(x - 5)^2– 3(x - 2)?

Mathematics

1 answer:

spayn [35]2 years ago

7 0

Answer:

4x² - 43x - 106

Step-by-step explanation:

4(x - 5)²– 3(x - 2)

4(x - 5)(x - 5) - 3 (x - 2)

distribute

4(x(x - 5) - 5(x - 5)) - 3 (x - 2)

4(x² - 5x - 5x - 25) - 3 (x - 2)

↓

combine like terms ↓

4(x² - 10x - 25) ((( ↓

distribute again ↓

4x² - 40x - 100 ↓

↓

distribute ↔ ↓

4x² - 40x - 100 - 3 (x - 2)

4x² - 40x - 100 - 3x + 6

add

4x² - 40x - 106 - 3x

combine

4x² - 43x - 106

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In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year. Let X be the number of quakes in a given

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a) Earthquakes are random and independent events.

b) There is an 85.71% probability of fewer than three quakes.

c) There is a 0.51% probability of more than five quakes.

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 time interval.

In this problem, we have that:

In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year, so $\mu = 1.3$

(a) Justify the use of the Poisson model.

Earthquakes are random and independent events.

You can't predict when a earthquake is going to happen, or how many are going to have in a year. It is an estimative.

(b) What is the probability of fewer than three quakes?

This is $P(X < 3)$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725$

$P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543$

$P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2725 + 0.3543 + 0.2303 = 0.8571$

There is an 85.71% probability of fewer than three quakes.

(c) What is the probability of more than five quakes?

This is $P(X > 5)$

We know that either there are 5 or less earthquakes, or there are more than 5 earthquakes. The sum of the probabilities of these events is decimal 1.

$P(X \leq 5) + P(X > 5) = 1$

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725$

$P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543$

$P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303$

$P(X = 3) = \frac{e^{-1.3}*1.3^{3}}{(3)!} = 0.0980$

$P(X = 4) = \frac{e^{-1.3}*1.3^{4}}{(4)!} = 0.0324$

$P(X = 5) = \frac{e^{-1.3}*1.3^{5}}{(5)!} = 0.0084$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2725 + 0.3543 + 0.2303 + 0.0980 + 0.0324 + 0.0084 = 0.9949$

Finally

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9949 = 0.0051$

There is a 0.51% probability of more than five quakes.

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