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4 years ago

I need help please. DUE TOMORROW!

Mathematics

2 answers:

Julli [10]4 years ago

5 0

In order to get your final answer to this problem,, you will need to factor out $\frac{2}{9}$ from the expression.
$\frac{2}{9}$ × (j + 3)
Since we cannot simplify this expression down any further and we have removed our coefficient,, the correct answer to your question is going to be $\frac{2}{9}$ × (j + 3).
Let me know if you have any further questions.
:)

amid [387]4 years ago

5 0

You are given (2/9)j + 2/3. The variable is j and the coefficient of j is (2/9). <em>If you're supposed to factor out the coeff. of j, factor out (2/9) from both (2.9)j and (2/3):

(2/9)(j + 3) (answer)
</em>

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Solve the equation <br> 14(3x-1)=2x-23

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$x=\frac{-9}{40} \\$

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3 years ago

A major hurricane is a hurricane with wind speeds of 111 miles per hour or greater. During the 20th century the mean number of t

Bas_tet [7]

Answer:

a) 0.3293 = 32.93% probability that the number of major hurricane to strike U.S mainland in any any give year is exactly one.

b) 0.878 = 87.8% probability that the number of major hurricane to strike U.S mainland in any any give year is at most one.

c) 0.122 = 12.2% probability that the number of major hurricane to strike U.S mainland in any any give year is more than one.

Step-by-step explanation:

We have the mean during an interval, which means that the Poisson distribution is used to solve this question.

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.

During the 20th century the mean number of the major hurricanes to strike U.S mainland per year was 0.6.

This means that $\mu = 0.6$

Find the probability that the number of major hurricane to strike U.S mainland in any any give year is:

a) Exactly one

This is P(X = 1). So

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

$P(X = 1) = \frac{e^{-0.6}*0.6^{1}}{(1)!} = 0.3293$

0.3293 = 32.93% probability that the number of major hurricane to strike U.S mainland in any any give year is exactly one.

b) At most one

This is:

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

Then

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

$P(X = 0) = \frac{e^{-0.6}*0.6^{0}}{(0)!} = 0.5488$

$P(X = 1) = \frac{e^{-0.6}*0.6^{1}}{(1)!} = 0.3293$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.5488 + 0.3292 = 0.878$

0.878 = 87.8% probability that the number of major hurricane to strike U.S mainland in any any give year is at most one.

c) More than one

This is:

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.878 = 0.122$

0.122 = 12.2% probability that the number of major hurricane to strike U.S mainland in any any give year is more than one.

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