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ddd [48]
3 years ago
12

OK this is another question which I need help on please please help if you can. CAUSED ME MUCH DISTRESS

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
3 0

Answer:

15 = 2x - 3y

Step-by-step explanation:

We have the two points P1(3,-3) and P2(9,1).

First, calculate the slope between those points:

slope = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-3)}{9 - 3} = \frac{4}{6} = \frac{2}{3}

Now simply move 3 steps from (3,-3) towards the y-axis to get the intersection with the y-axis (each step reduces x by 1 and applies the slope to the y-value):

(3,-3)

---> (2, -3 - (2/3)) = (2,-3 2/3)

---> (1, (-3 2/3) - 2/3) = (1, -4 1/3)

---> (0, (-4 1/3) - 2/3) = (0,-5)

This tells us, that our line intersects the y-axis at (0,-5).

The general form of a line is f(x) = <SLOPE> * x + <Y-VALUEATXEQUALSZERO>.

In our case: f(x) = y = (2/3)x -5.

To get it into the correct form, we simply subtract y from both sides and get:

0 = (2/3)x - y - 5

To get only integers just multiply everything with 3 and add 15 and get:

15 = 2x - 3y, which fulfils the form asked for in the problem.

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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:

180-108=72

where I got 180: a supplementary angle is 180 degrees

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