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raketka [301]
3 years ago
13

Which of the following equationa apply to line below

Mathematics
2 answers:
jeka943 years ago
3 0

Answer:

1+1=magelan

2plus2 lapulapu

Georgia [21]3 years ago
3 0

Answer:

b and d

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

m is the slope and (a, b) a point on the line

Calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (- 3, 6) and (x₂, y₂ ) = (- 2, 2) ← 2 points on the line

m = \frac{2-6}{-2+3} = - 4

Using (a, b ) = (- 3, 6 ), then

y - 6 = - 4(x - (- 3)), that is

y - 6 = - 4(x + 3) → d

Using (a, b ) = (- 2, 2 ), then

y - 2 = - 4(x - (- 2)) , that is

y - 2 = - 4(x + 2) →b

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3 years ago
An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.
andrey2020 [161]

Answer:

The probability that A selects the first red ball is 0.5833.

Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are E(1)= ^9C_2 places in which the other 2 red balls can be placed.

A red ball drawn third, there are E(3)= ^7C_2 places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are E(5)= ^5C_2 places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are E(7)= ^3C_2 places in which the other 2 red balls can be placed.

The total number of total event is S= ^{10}C_3

The probability that A selects the first red ball is

P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}

P(A \text{wins})=\frac{36+21+10+3}{120}

P(A \text{wins})=\frac{70}{120}

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6 0
3 years ago
I have some questions regarding combinations in finite. Here is my first one I am stumped on:
podryga [215]

We will get the number of possible selections, and then subtract the number less than 25 cents.

We can choose the number of dimes 5 ways 0,1,2,3 or 4.
We can choose the number of nickels 4 ways 0,1,2 or 3.
We can choose the number of quarters 3 ways 0,1, or 2.

That's 5*4*3  = 60 selections 

Now we must subtract from the 60 the number of selections of coins that are less than 25 cents. These will involve only dimes and nickels. 

To get a selection of coin worth less than 25 cents:
If we use no dimes, we can use 0,1,2  on all 3 nickels.
That's 4 selections less than 25 cents. (that includes the choice of No coins at all in the 60, which we must subtract).

If we use exactly 1 dime , we can use 0,1,2, or all 3 nickels.
That's the 3 combinations less than 25 cents.

And there is 1 other selection less than 25 cents, 2 dimes and no nickels. 

So that's 4+3+1 = 8 selections which we must subtract from the 60.
 
Answer 60-8 = 52 selections of coins worth 25 cents or more.

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