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Ostrovityanka [42]

3 years ago

8

Caroline, Krutika and Natasha share some sweets in the ratio 4:3:3. Caroline gets 9 more sweets than Natasha. How many sweets do

es Krutika get?

1 answer:

vagabundo [1.1K]3 years ago

3 0

Answer:

Krutika gets 27 sweets

Step-by-step explanation:

Let

x ---> number of sweets that Caroline gets

y ---> number of sweets that Krutika gets

z ---> number of sweets that Natasha gets

we know that

$x:y:z=4:3:3$

so

$\frac{x}{y}=\frac{4}{3}$

$x=\frac{4}{3}y$ ----> equation A

$\frac{x}{z}=\frac{4}{3}$

$x=\frac{4}{3}z$ ----> equation B

$y=z$ ---> equation C

Caroline gets 9 more sweets than Natasha

so

$x=z+9$ ---> equation D

substitute equation D in equation B

$z+9=\frac{4}{3}z$

solve for z

$\frac{4}{3}z-z=9$

$\frac{1}{3}z=9\\z=27\ sweets$

therefore

Krutika gets 27 sweets

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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 $P(X \geq 4)$ .

Either Florida is struck by less than four hurricanes in a given year, or it is struck by 4 or more. The sum of these probabilities is decimal 1.

$P(X < 4) + P(X \geq 4) = 1$

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

So

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

$P(X = 0) = \frac{e^{-0.704}*(0.704)^{0}}{(0)!} = 0.4946$

$P(X = 1) = \frac{e^{-0.704}*(0.704)^{1}}{(1)!} = 0.3482$

$P(X = 2) = \frac{e^{-0.704}*(0.704)^{2}}{(2)!} = 0.1226$

$P(X = 3) = \frac{e^{-0.704}*(0.704)^{3}}{(3)!} = 0.0288$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.4946 + 0.3482 + 0.1226 + 0.0288 = 0.9942$

Finally

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.9942 = 0.0058$

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

7 0

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