Jake draws A POR on the coordinate plane

a highschool computer club has 15 members . how many different ways can they choose a president and vice president

Nataliya [291]

Answer=210 ways

Okay. I'm not entirely sure this is the BEST way to solve for this answer, but nonetheless, I'll show how I would solve the equation.

First, lets look at the probability of the two options happening. The probability of one of the 15 members becoming president is 1/15. Then, since one student is already being used for the spot of president, the odds of another student becoming vice president is 1/14.

$\frac{1}{15} * \frac{1}{14} = \frac{1}{210}$

From this, we know that there are 210 options for the positions of president and vice president total (since 210 represents the whole).

So they can choose the president and vice president 210 different ways

3 0

3 years ago

1. (2,7); m=-4 What’s the answer?

Nataly_w [17]

Answer:

4x + y = 15 or y = -4x + 15

Step-by-step explanation:

7 = 2[-4] + b

-8

15 = b

$y = -4x + 15$

If you want it in Standard Form:

y = -4x + 15

+4x +4x

___________

4x + y = 15 >> Line in Standard Form

I am joyous to assist you anytime.

4 0

3 years ago

Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea

elena-s [515]

Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) = P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725

There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

$nCk=\frac{n!}{k!(n-k)!}$

So, the number of ways to select exactly 3 aces is:

$4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192$

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

$P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}$

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

$P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052$

5 0

3 years ago

A public interest group hires students to solicit donations by telephone. After a brief training period students make calls to p

IceJOKER [234]

I need help with the answer?

8 0

3 years ago

Y=1/4x−2y=−2x+3 help solve problem.

8_murik_8 [283]

Answer:

(x, y) = (2 2/9, -1 4/9)

Step-by-step explanation:

Equate the values of y and solve for x.

1/4x -2 = -2x +3

(2 1/4)x = 5 . . . . . . . . add 2+2x to both sides

x = 20/9 = 2 2/9 . . . multiply by 4/9

y = -2(2 2/9) +3 = -4 4/9 +3 . . . . substitute for x in the second equation

y = -1 4/9

The solution is x = 2 2/9, y = -1 4/9.

7 0

2 years ago