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bixtya [17]
2 years ago
7

A missile protection system consists of n radar sets operating independently, each with probability 0.9 of detecting a missile e

ntering a zone that is covered by all of the units.a) Find a nice expression for the probability of detecting at least one of the sets detects the missile. (Tip: what is the probability, that a missile is not detected by any of the sets? - and what has this probabilityto do with the above?)b) How large must n be if we require that the probability of detecting a missile that enters the zone be 0.999 ? 0.999999 ?c) If n = 5 and a missile enters the zone, what is the probability that exactly 4 sets detect the missile?At least one set?
Mathematics
1 answer:
stepladder [879]2 years ago
6 0

Answer:

A.) The probability that the missile is not detected by any of n sets = (1-0.9)^n = 0.1^n.

Where it's opposite will be that the missile is detected by atleast 1 of the sets. Which will be equals to 1-(0.1)^n.

B.) 0.999= 1-(0.1)^n

Here n=3

For, 0.999999=1-(0.1)^n

n=6.

C.) If n=5

5C4*(0.9)^4*(0.1)^1= 0.32805

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15000

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8 0
2 years ago
A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and
natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: E=k\frac{q}{r^{2} }; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: p=\frac{q}{A}; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: S=4\pi r^{2}; which gives us the surface of a sphere.

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Then, we are able to calculate both fields:

Inner field: Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}

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The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

3 0
3 years ago
How would I solve <br> -6.11 + b = 14.321
Studentka2010 [4]

Answer:

b=20.431

Step-by-step explanation:

you just add  6.11 to both sides. on the left side the -6.11 and +6.11 cancel out and on the right side you will just add 6.11 to 14.321. when you add 14.321 and 6.11 you get 20.431. therefore, b=20.431

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