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puteri [66]
3 years ago
10

Find the simple interest on RS.2500 for 2 years and 6 months at 3% per annum.Also find the amount.

Mathematics
1 answer:
meriva3 years ago
6 0

187.50

simple interest (I ) = PRT

where P is the Principle ( amount deposited ), R is the rate of interest and T the time in years

here P = 2500, R = 3% and T = 2.5 years

I = 2500 × \frac{3}{100} × 2.5

 = 25 × 3 × 2.5 = 187.50


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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.

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Question

Cable company A charges $45 a month for cable plus a $18 installation fee. Cable B charges $39 a month for cable plus a $30 fee for installation. Which inequality can be used to find out when the monthly cost for Cable company A is less than Cable company B?

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Step-by-step explanation:

Step 1

We have to find the Algebraic expressions for the cable companies

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Hence, this is represented as:

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