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mars1129 [50]
3 years ago
12

Which point is on the line that passes through point R and is perpendicular to line PQ? (–6, 10) (–4, –8) (0, –1) (2, 4)

Mathematics
2 answers:
Eddi Din [679]3 years ago
5 0

Answer:

It's B i did the unit test review thing

Step-by-step explanation:

Paraphin [41]3 years ago
4 0

Answer:

(-4, -8)

Step-by-step explanation:

Let the coordinates of common point of the given lines are (x, y),

Thus, the slope of the line passes through the points (a, b) and R(4,2) is,

       2 - b

m1 = ------

        4 - a

Again, the slope of the line passes through two points P(-6,4) and Q(4,-4),

      -4 - 4          -8           -8          -4

m2 = ------  =   --------- = --------- = ---------

        4- (-6)       4 + 6         10        5

= m1 * m2 = -1

 

  2 - b      -4

= -------   x ---- =  -1

   4 - a      5

        8 - 4b

=     -----------   = 1

        20 - 5a

= 8 - 4b  = 20 - 5a

= 5a - 4b = 12 --------- equation 1

<u>For a = -6   and b = -10</u>

5 x -6  - 4  x  10 = -70 ≠ 12

<u>For a = -4   and b = -8</u>

5 x -4  - 4  x  - 8  = 12 = 12

<u>For a = 0  and b = -1</u>

5 x 0 - 4 x -1 = 4 ≠ 12

<u>For a = 2 and b = 4</u>

5 x 2 - 4 x 4 = -6 ≠ 12

therefore Second is correct

hope it helps and i get the brainliest.

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                                        HOPE THIS HELPED!!!

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3 years ago
4. Assume that the chances of a basketball player hitting a 3-pointer shot is 0.4 and the probability of hitting a free-throw is
vovikov84 [41]

Answer:

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

Step-by-step explanation:

For each 3-pointer shot, there are only two possible outcomes. Either the player makes it, or the player does not. The same is valid for free throws. This means that both the number of 3-pointers and free throws made are given by binomial distributions.

Since 3-pointers and free throws are independent, first we find the probability of making exactly 3 3-pointers out of 10, then the probability of making exactly 5 free throws out of 10, and then the probability that the player will make exactly 3 3-pointers and 5-free throws is the multiplication of these probabilities.

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 combinations 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.

Probability of making 3 3-pointers out of 10:

The chances of a basketball player hitting a 3-pointer shot is 0.4, which means that p = 0.4. So this is P(X = 3) when n = 10.

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

P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.21499

Probability of making 5 free throws out of 10:

The probability of hitting a free-throw is 0.65, which means that p = 0.65. The probability is P(X = 5) when n = 10.

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

P(X = 3) = C_{10,5}.(0.65)^{5}.(0.35)^{5} = 0.15357

Calculate the probability that the player will make exactly 3 3-pointers and 5-free throws.

0.21499*0.15537 = 0.0334

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

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Answer:

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Step-by-step explanation:

Is this what you meant?  If so heres' the answer

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