Given GHI G(4,-3), H(-4,2), and I(2,4), find the perpendicular bisector of HI in standard form.

1 answer:

8 0

Finding the slope of HI ( $m$ ):

$m=\dfrac{\Delta y}{\Delta x}\iff m=\dfrac{y_I-y_H}{x_I-x_H}=\dfrac{4-2}{2-(-4)}\iff m=\dfrac{2}{6}\iff\\\\\boxed{m=\dfrac{1}{3}}$

Finding the perpendicular slope ( $m'$ ):

$m\cdot m'=-1\iff \dfrac{1}{3}m'=-1\iff\boxed{m'=-3}$

Using the formula:

$y-y_G=m'(x-x_G)\iff y-(-3)=-3(x-4)\iff\\\\ y+3=-3x+12\iff y=-3x+9$

Rewriting:

$\boxed{y+3x=9}$

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An unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6

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Answer:

Probability of rolling at least 4 sixes is 0.01696.

Step-by-step explanation:

We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6 times.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......$

where, n = number trials (samples) taken = 6 trials

r = number of success = at least 4

p = probability of success which in our question is probability of

rolling a “six", i.e; p = 0.20

Let X = Number of sixes on a die

So, X ~ Binom(n = 6, p = 0.20)

Now, Probability of rolling at least 4 sixes is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = P(X = 4) + P(X = 5) + P(X = 6)

= $\binom{6}{4} \times 0.20^{4} \times (1-0.20)^{6-4}+\binom{6}{5} \times 0.20^{5} \times (1-0.20)^{6-5}+\binom{6}{6} \times 0.20^{6} \times (1-0.20)^{6-6}$