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netineya [11]
3 years ago
12

Find the distance between the points (0,2)and (-5,-9)The answer must be a whole number or a fully simplified radical expression

do not round up
Mathematics
1 answer:
LiRa [457]3 years ago
7 0

Answer:

d = \sqrt{146}

Step-by-step explanation:

d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d= \sqrt{(-5-0)^2+(-9-2)^2}\\d= \sqrt{(-5)^2+(-11)^2}\\d= \sqrt{25+121}\\d= \sqrt{146}

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Write an equation in slope-intercept form of the line through point P(6, –1) with slope 4.
umka21 [38]
Y = mx+b

m= 4 

y= 4x+b

solve for b using (6,-1)

-1 = 4 (-1)+b
b=3

therefore equation is y=4x+3

same as (y+1)=4(x-6)
3 0
3 years ago
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How many time can 2/3 go into 20
kramer

Answer: 30 times

2/3=.66...

20/.66=30.30

I hope this helps:

6 0
3 years ago
Which equation represents the polar form of x2 + (y + 4)2 = 16?
True [87]

Answer:

r = -8sin(theta)

Step-by-step explanation:

used desmos

6 0
2 years ago
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A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit
Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution  

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

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

The parameters are:

  • p is the probability of a success on a single trial.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

The expected number of <u>trials before q successes</u> is given by:

E = \frac{q(1-p)}{p}

The variance is:

V = \frac{q(1-p)}{p^2}

In this problem, 0.2 probability of a finding a person who attended the last football game, thus p = 0.2.

Item a:

  • None of the first three attended, which is P(X = 0) when n = 3.
  • Fourth attended, with 0.2 probability.

Thus:

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

P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512

0.2(0.512) = 0.1024

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

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

P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c:

  • One person, thus q = 1.

The expected value is:

E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4

The variance is:

V = \frac{0.8}{0.04} = 20

The expected number of people is 4, with a variance of 20.

A similar problem is given at brainly.com/question/24756209

3 0
1 year ago
12 cm
Ahat [919]

Answer:

The answer is 200 cm³

Step-by-step explanation:

4 0
2 years ago
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