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4 years ago

Tina wants to decorate a rectangular picture frame measuring 18 w 14 tall she has 4 ft of rope dies she have enough rope

Mathematics

1 answer:

igor_vitrenko [27]4 years ago

5 0

Tina does not have enough rope to decorate the frame

Step-by-step explanation:

Tina wants to decorate a rectangular picture frame

The width of the frame is 18 inches
The length of the frame is 14 inches
She has 4 feet rope

We want to know if she has enough rope or not to decorate the frame

∵ The frame is in the shape of the rectangle

- The length of the rope must be equal the perimeter of the frame

∵ The perimeter of a rectangle = 2 length + 2 width

∵ The length = 14 inches

∵ The width = 18 inches

∴ The perimeter of the frame = 2(14) + 2(18) = 28 + 36 = 64 inches

∵ The length of the rope is 4 feet

∵ 1 foot = 12 inches

∴ The length of the rope = 4 × 12 = 48 inches

∵ 48 < 64

∴ The length of the rope is less than the perimeter of the frame

∴ Tina does not have enough rope

Tina does not have enough rope to decorate the frame

Learn more:

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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

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

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

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

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

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

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

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

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.

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3 years ago

Louis wants to give \$15 to help kids who need school supplies. He also wants to buy a pair of shoes for \$39.

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Answer:

Louis needs to save 15+39 = $ 54 dollars

Step-by-step explanation:

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DochEvi [55]

The current Brainliest answer seems to be answering the question "Every integer is a multiple of which number?" rather than the question presented here.

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By 2's: 0, 2, 4, 6, 8

By 6's: 0, 6, 12, 18

By 7's: 0, 7, 14, 21

Because we can always "reach" 0 regardless of the integer we're counting by, we can say that 0 is a multiple of every integer.

More formally, we say that some number n is a multiple of an integer x if we can find another integer y so that x · y = n. By this definition, 18 would be a multiple of 6 because 6 · 3 = 18, and 3 is an integer. We can use the property that the product of any number and 0 is 0 to say that x · 0 = 0, where x can be any integer we want. Since 0 is also an integer, this means that, by definition, 0 is a multiple of every integer.

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