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zhuklara [117]
3 years ago
5

Military radar and missile detection systems are designed to warn a country of enemy attacks. A reliability question deals with

the ability of the detection system to identify an attack and issue the warning. Assume that a particular detection system has a 0.83 probability of detecting a missile attack. Answer the following questions using the binomial probability distribution:
a. What is the probability that a single detection system will detect an attack?
b. If two detection systems are installed in the same area and operate independently, what
is the probability that at least one of the systems will detect the attack?
c. If three systems are installed, what is the probability that at least one of the systems
will detect the attack?
d. Would you recommend that multiple detection systems be used? Explain.
Mathematics
1 answer:
Nadusha1986 [10]3 years ago
7 0

Answer:

Step-by-step explanation:

Given that a particular detection system has a 0.83 probability of detecting a missile attack.

Also each system is independent of the other to detect missile attack.

a)  the probability that a single detection system will detect an attack=0.83

b) the probability that at least one of the systems will detect the attack given two systems= 1-Prob no system detects

=1-(1-0.83)(1-0.83)\\\\=1-0.0289\\=0.9711

c)  If three systems are installed,  the probability that at least one of the systems  will detect the attack

= 1- none of the three systems detect attack

= (1-0.17^3)\\=1-0.004913\\=0.995087

d) We find that probability of detecting is more if we increase the system form 1 to 2 and 2 to 3.

Hence for safety and defence purpose multiple detection system is recommended.

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Can someone please solve these?
JulijaS [17]

Answer:

Step-by-step explanation:

h = -6t^{2} +20t + 4

I will use calculus,  maybe that's not how you're supposed to do this

-12t +20 =0

12t = 20

t = 20 /12

t = 1 \frac{8}{12}

t = 1 \frac{2}{3}

there will be a max at  1.6666666666  seconds

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= 16.666666666666 + 33.333333333333 + 4

= - 16\frac{2}{3} + 33 \frac{1}{3} + 4

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time of flight:

0 = -6t^{2} +20t + 4

use quadratic formula to find t

-20 +- sqrt [ 20^{2} - 4*(-6)*4 ] / 2*(-6)

-20 +- sqrt [400 + 96 ] / -12

-20 +- sqrt [496 ] / -12

-20 +- 22.27105 / -12

try the negative option 1st

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when will the rocket be at 12' ? :

12 = -6t^{2} +20t + 4

0 = -6t^{2} +20t -8

use quadratic formula again to find t

-20 +- sqrt [ 20^{2} - 4*(-6)*(-8) ] / 2*(-6)

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-20 +- sqrt [208 ] / -12

-20 - 14.4222 / -12

-34.4222 / -12

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and

-20 + 14.4222 / -12

-5.578 / - 12

0.4648  seconds ( on the way up )

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Move constant to the right side and change its sign

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Divide both sides of the equation by -3

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