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3 years ago

Help me i cant figure this out

Mathematics

1 answer:

scoray [572]3 years ago

8 0

McKenzie's work is correct because on step 3 she added the A and the B correctly.

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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

viva [34]

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:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

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

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

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.

8 0

4 years ago

A chemist is using 362 milliliters of a solution of acid and water. If 12.1% of the solution is acid, how many milliliters of ac

PSYCHO15rus [73]

Answer:

The answer to your question is 43.8 mL of acid.

Step-by-step explanation:

Data

Total volume = 362 ml

12.1 % is acid

The Volume of acid = ?

Process

1.- Use the rule of three and cross multiplication to solve this problem.

362 ml ------------------- 100%

x ------------------- 12.1 %

x = (12.1 x 362) / 100

x = 4380.2 / 100

x = 43.802 ml

2.- Round your answer to the nearest tenth.

43.802 only consider the first decimal

43.8 mL

3.- Conclusion

A 12.1% solution of acid has 43.8 mL of acid.

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25% of ____ is 100? Help meh.

levacccp [35]

400

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Which is equivalent to 38*?<br> o

xxMikexx [17]

8³x

$\sqrt[3]{83}$

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4 years ago