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3 years ago

Please answer this question please

Mathematics

1 answer:

Bas_tet [7]3 years ago

8 0

<h3>Answer:</h3><h3 /><h3>Amount = Rs 13891.50</h3><h3>Compound interest = Rs 1891.50</h3>

Step-by-step explanation:

C = Amount (A) - Principal (P)

Where

C is the compound interest

To find the amount we use the formula

$A = P ({1 + \frac{r}{100} })^{n}$

where

P is the principal

r is the rate

n is the period / time

From the question

P = Rs 12, 000

r = 5%

n = 3 years

Substitute the values into the above formula

That's

$A = 12000 ({1 + \frac{5}{100} })^{3} \\ A = 12000(1 + 0.05)^{3} \\ A = 12000 ({1.05})^{3}$

We have the answer as

<h3>Amount = Rs 13891.50</h3>

Compound interest = 13891.50 - 12000

<h3>Compound interest = Rs 1891.50</h3>

Hope this helps you

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Answer:

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For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

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In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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And p is the probability of X happening.

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The probability that a single radar station will detect an enemy plane is 0.65. This means that $n = 0.65$ .

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which $P(X = 0) \leq 0.02$ .

n = 1.

$P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35$

n = 2

$P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225$

n = 3

$P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429$

n = 4

$P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015$

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(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

$E(x) = np$

So when $n = 7$

$E(x) = 7*(0.65) = 4.55$

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

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