11b+6 >/=14b+3 please help

Mathematics

1 answer:

Mazyrski [523]3 years ago

4 0

Answer:

Solving the inequality $11b+6 >=14b+3$ we get b <= 1

Step-by-step explanation:

We need to solve the inequality $11b+6 >=14b+3$

solving the inequality:

$11b+6 >=14b+3$

Subtracting 6 on both sides

$11b+6-6 >=14b+3-6\\11b >= 14b-3$

subtracting 14 b on both sides

$11b-14b >= 14b-3-14b\\-3b >= -3$

Divide both sides by -3 and the inequality will be changed to less than equal i.e <=

$\frac{-3b}{-3}$

So, solving the inequality $11b+6 >=14b+3$ we get b <= 1

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Since 1851, exactly 119 hurricanes have hit Florida (this includes the years 1851 and 2019),—only direct hits by hurricanes are

alukav5142 [94]

Answer:

There is a 0.58% probability of Florida being struck by four or more hurricanes in the same year.

Step-by-step explanation:

Since we have only the mean during the interval, we can solve this problem using the Poisson probability distribution.

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

Poisson probability distribution

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

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:

Since 1851, exactly 119 hurricanes have hit Florida (this includes the years 1851 and 2019). Counting 1851 and 2019, there are 169 years in this interval. This means that $\mu = \frac{119}{169} = 0.704$

If the probability of hurricane strikes has remained the same since 1851, what is the probability of Florida being struck by four or more hurricanes in the same year?

This is $P(X \geq 4)$ .