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velikii [3]
3 years ago
7

Triangle ABC has vertices A(0,6) B(4,6) C(1,3). Sketch a graph of ABC and use it to find the orthocenter of ABC. Then list the s

teps you took to find the orthocenter, including any necessary points or slope you had to derive.
Mathematics
1 answer:
Allisa [31]3 years ago
5 0

For line B to AC:  y - 6 = (1/3)(x - 4);  y - 6 = (x/3) - (4/3); 3y - 18 = x - 4, so 3y - x = 14

For line A to BC:  y - 6 = (-1)(x - 0);  y - 6 = -x, so y + x = 6

Since these lines intersect at one point (the orthocenter), we can use simultaneous equations to solve for x and/or y:

(3y - x = 14) + (y + x = 6) =>  4y = 20, y = +5;  Substitute this into y + x = 6:  5 + x = 6, x = +1

<span>So the orthocenter is at coordinates (1,5), and the slopes of all three orthocenter lines are above.</span>

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. There are 11 boys and 5 girls that tried out to
Lisa [10]

We want to find the probability that the two students chosen for the duet are boys. We will find that the probability that both students chosen for the duet are boys is 0.458

If we assume that the selection is totally random, then all the students have the same<em> </em><em>probability </em><em>of being chosen.</em>

This means that, for the first place in the duet, the probability of randomly selecting a boy is equal to the quotient between the number of boys and the total number of students, this is:

P = 11/16

For the second member of the duet we compute the probability in the same way, but this time there is one student less and one boy less (because one was already selected).

Q = 10/15

The joint probability (so both of these events happen together) is just the product of the individual probabilities, this will give:

Probability = P*Q = (11/16)*(10/15) = 0.458

So the probability that both students chosen for the duet are boys is 0.458

If you want to learn more, you can read:

brainly.com/question/1349408

5 0
2 years ago
Which is the correct graph of y=2/x^2-4 ?
Elis [28]
The correct graph would be the first one

6 0
3 years ago
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
2 years ago
Find the value of x. No calculator needed. Leave the square root as a part of your answer.
aleksklad [387]

Answer:

5/V2=2.5×V2

Step-by-step explanation:

This is an isoscelle right triangle:

x²+x²=5²

2x²=5²

x=5/V2=5×V2/2=2.5×V2

where V2=sqrt(2)

4 0
3 years ago
Paul has $20,000 to invest. His intent is to earn 10.5% interest on his investment. He can invest part of his money at 7% intere
Art [367]
X=7%
y=12%
x+y=20000
0.07x+0.12y=20000*0.19=3800
3 0
3 years ago
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