Graph the equation. y-4=-5x(x-1)

Travka [436]

4(x+x+7)-2x+8-4
Substitute x with 1
4(1+1+7)-2(1)+8-4
=4(2+7)-2+4
=4(9)-2+4
=36-2+4
=34+4
=38

4(x+x+7)-2x+8-4
Substitute x with 2
4(2+2+7)-2(2)+8-4
=4(4+7)-4+4
=4(11)
=44. Hope it help!

3 0

3 years ago

Joianie beleives that you cannot use cross products to solve the proportion 50/3 = 75/x for x. She says

BigorU [14]

Answer:

4.5 It doesn't come out evenly, but cross products do work.

Step-by-step explanation:

She is not correct.

50 * x = 3 * 75 Combine the right

50x = 225 Divide by 50

x = 225/50

x = 4.5

6 0

3 years ago

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 $P(X \geq 4)$ .