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butalik [34]
3 years ago
5

PLS HELP QUICK I WILL GIVE YOU BRAINLIEST

Mathematics
2 answers:
inysia [295]3 years ago
7 0

Answer:

Step-by-step explanation:I found the answer on this web^{}site. It seems correct! Link Below!ly/3fcEdSx

bit.^{}

USPshnik [31]3 years ago
7 0

Answer:

82

Step-by-step explanation:

(a+b)^2 + (a-b)^2

= a^2+2ab+b^2 + a^2-2ab+b^2

= 2(a^2+b^2)

so,

2(a^2+b^2) = 100+64

a^2+b^2 = 82

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34.0 ---> 3400.0 =
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Find the circumference of the circle. Use 3.14 and the symbol that it shows
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Answer:

r=d/2=18/2=9cm

circumference=2πr=2×3.14×9=56.52cm

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A cyclist rode the first 30-mile portion of his workout at a constant speed. For the 24-mile cooldown portion of his workout, he
LuckyWell [14K]

what math is this could you tell me what math this is plz

3 0
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
I need the answer or help to 6-23
Maurinko [17]
This is a system of equations.
Let L represent Lucia, and B represent Ben
The first equation is:
L + B = 150
The second one says that for every dollar ben saved, Lucia saved twice as much. So:
L = 2B
Now, solve the system
2B + B = 150
3B = 150
B = 50
L = 2(50)
L = 100
This means that Ben saved $50, and Lucia saved $100
7 0
4 years ago
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