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3 years ago

Help with algebra 1 (A-y)^2 +2y^3 If a=3 and y=-2

Mathematics

1 answer:

zlopas [31]3 years ago

4 0

Answer:

9

Step-by-step explanation:

(a - y)^2 + 2y^3 where a = 3 and y = -2.

Step 1. Substitute a for 3 and y for -2

((3) - (-2))^2 + 2(-2)^3

Step 2. Subtract

(5)^2 + 2(-8)

Step 3. Simplify the exponent

25 + 2(-8)

Step 4. Multiply

25 + (-16)

Step 5. Add

9
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Hope this helps!

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- doomdabomb

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The mean number of automobiles entering a mountain tunnel per two-minute period is one. An excessive number of cars entering the

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Answer:

A Poisson model seems reasonable for this problem, since we have the mean during the time interval.

There is a 1.9% probability that the number of autos entering the tunnel during a two-minute period exceeds three.

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:

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In which

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This means that $\mu = 1$ .

For a Poisson model to be reasonable, we only need the mean during the time interval. So yes, a Poisson model seems reasonable for this problem.

Find the probability that the number of autos entering the tunnel during a two-minute period exceeds three.

We want to find $P(X>3)$

Either this number is less or equal to 3, or it exceeds 3. The sum of the probabilities is decimal 1. So:

$P(X \leq 3) + P(X > 3) = 1$

$P(X > 3) = 1 - P(X \leq 3)$

In which

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

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

$P(X = 0) = \frac{e^{-1}*1^{0}}{(0)!} = 0.3679$

$P(X = 1) = \frac{e^{-1}*1^{1}}{(1)!} = 0.3679$

$P(X = 2) = \frac{e^{-1}*1^{2}}{(2)!} = 0.1839$

$P(X = 3) = \frac{e^{-1}*1^{3}}{(3)!} = 0.0613$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3679 + 0.3679 + 0.1839 + 0.0613 = 0.981$

Finally

$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.981 = 0.019$

There is a 1.9% probability that the number of autos entering the tunnel during a two-minute period exceeds three.

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