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3 years ago

Please say the answer as fast possible with working. It would be so kind if you sended this early

Mathematics

1 answer:

Nimfa-mama [501]3 years ago

5 0

Answer: 2. 2

3. 110,121,132

4.22,110

5.4.5

Step-by-step explanation:

2. 20-12=8, 8divided by 4=2

3. 363 divided by 3=121(this is the avg of the consecutive multiples so one multiple must be 11 above and another 11 below

4.66 divided by 3=22, 22x 2 and 22 x 5

5.2 x2=4, 13-4=9, 9 divided by 2 is 4.5

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Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

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

Step-by-step explanation:

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Binomial probability distribution

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$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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

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So when $n = 7$

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