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3 years ago

7

Vanessa draws one side of an equilateral triangle ABC on the coordinate plane at points A(-2, 1) and B (4, 1). What are the two

possible coordinates of vertex C? Round to the nearest tenth.

1 answer:

Readme [11.4K]3 years ago

5 0

Answer:

two vertex are: (1 , 6.2) , (1 , - 4.2)

Step-by-step explanation:

A(-2, 1) and B (4, 1) AB is a parallel segment with x axis

The 3rd vertex P (x , y) will locate at the mid point of AB

x = (-2) + (4 - (-2)) /2 = 1 P (1 , y)

length AB = 4 - (-2) = 6

AP = AB (equilateral)

BP² = (4 - 1)² + (1 - y)² = 6²

1 - 2y + y² = 36 - 9 = 27

y² - 2y -26 = 0

y = ((-(-2)) ± √(-2)² - 4 x (-26)) / 2 = (2 ± √108) / 2 = 1 ± 5.2

y = 6.2 or y = -4.2

P (1 , 6.2) or P ( 1 , -4.2)

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Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

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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.

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There are 10 containers, so $n = 10$ .

A food-packaging apparatus underfills 10% of the containers, so $p = 0.1$ .

a) This is P(X = 1)

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

b) This is P(X = 3)

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

c) This is P(X = 9)

$P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000$

d) This is $P(X \geq 5)$ .

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

$P(X < 5) + P(X \geq 5) = 1$

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868$

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

$P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937$

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

$P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742$

So

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837$

Finally

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163$

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