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How can you use scale models to solve problems

tino4ka555 [31]

Scale drawings make it easy to see large things on paper, like a house or road

6 0

3 years ago

The offices of president, vice president, secretary, and treasurer for an environmental club will be filled from a pool of 15c

Alexus [3.1K]

Answer:

a) 5.13% probability that all of the offices are filled by members of the debate team

b) 2.56% probability that none of the offices are filled by members of the debate team

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the people are chosen is important, since the first person is the president, the second is the vice president and so on... So we use the permutations formula to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

(a) What is the probability that all of the offices are filled by members of the debate team?

Desired outcomes:

4(president, vice president, secretary, and treasurer) offices from a set of 8(candidates of the members debate team).

So

$D = P_{(8,4)} = \frac{8!}{(8-4)!} = 1680$

Total outcomes:

4 offices from a set of 15 candidates.

$T = P_{(15,4)} = \frac{15!}{(15-4)!} = 32760$

Probability:

$P = \frac{D}{T} = \frac{1680}{32760} = 0.0513$

5.13% probability that all of the offices are filled by members of the debate team

(b) What is the probability that none of the offices are filled by members of the debate team?

Desired outcomes:

4(president, vice president, secretary, and treasurer) offices from a set of 7(candidates who are not of the members debate team).

$D = P_{(7,4)} = \frac{7!}{(7-4)!} = 840$

Total outcomes:

4 offices from a set of 15 candidates.

$T = P_{(15,4)} = \frac{15!}{(15-4)!} = 32760$

Probability:

$P = \frac{D}{T} = \frac{840}{32760} = 0.0256$

2.56% probability that none of the offices are filled by members of the debate team

7 0

3 years ago

find the probability of picking a red marble and a green marble when 2 marbles are picked (without replacement)from a bag contai

mina [271]

Answer:

p = 6/11.

Step-by-step explanation:

So we have a bag that contains 6 red marbles and 6 green marbles.

Then the total number of marbles that are in that bag is:

6 + 6 = 12

There are 12 marbles in the bag, and we assume that all marbles have the same probability of being randomly drawn.

Now we draw two marbles, we want to find the probability that one is red and the other is green.

The first marble that we draw does not matter, as we just want the second marble to be of the other color.

So, suppose we draw a green one in the first attempt.

Then in the second draw, we need to get a red one.

The probability of drawing a red one will be equal to the quotient between the red marbles in the bag (6) and the total number of marbles in the bag (12 - 1 = 11, because one green marble was drawn already)

Then the probability is:

p = 6/11.

Notice that would be the exact same case if the first marble was red.

Then we can conclude that the probability of getting two marbles of different colors is:

p = 6/11.

6 0

3 years ago

1/4y+10=1/7y Solve for y

neonofarm [45]

Can u help me with my question no one will

3 0

3 years ago

Read 2 more answers

What is the order pairs for (-4,7) , (-6,-4) and (15,8) , (-17,9)

GuDViN [60]

(-4,7) , (-6,-4) and (15,8) , (-17,9)

4 0

2 years ago