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Dmitry_Shevchenko [17]
3 years ago
12

On number 19 this needs to be shown full work I don’t know where to start can someone help me now ASAP!!!

Mathematics
1 answer:
DENIUS [597]3 years ago
7 0

Answer:

1

sorry if I'm wrong, I'm in the 7th grade lol, did my best :) hope I sortof helped

Step-by-step explanation:

5-2=3

square root of 3 with 3 is 1

5 - 2 = 1 \\ 3 \sqrt{3 = 1}

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3 years ago
The probability that a single radar station will detect an enemy plane is 0.65.
taurus [48]

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:

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.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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.

In this problem, we have that:

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

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, 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.

6 0
3 years ago
On a subway route, station C is located at the midpoint between stations A and D. Station B is located at the midpoint between s
Finger [1]
The distance between C and A/D is 1,2 km. The distance between B and A/C is 0,6. It's in this order: A 0,6 km B 0,6 km C 1,2km D so the distance between B and D is 0,6 + 1,2 = 1,8 km
6 0
3 years ago
Read 2 more answers
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