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3 years ago

Decide which of the following is a reasonable answer for .

Mathematics

1 answer:

Nikolay [14]3 years ago

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Follow for what REEEEE

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Mark is going to an awards dinner and wants to dress appropriately. He had one blue shirt, one white dress shirt, one black dres

lidiya [134]

Answer:

<u>All the four statements are true , which are -</u>

(A) The subset contains contains of all the outfits that have grey slacks .

(B) The subset contains contains of all the outfits that have grey slacks and red tie .

(C) The subset contains contains of all the outfits that have grey slacks and blue shirt .

(D) The subset contains contains of all the outfits that do not have black slacks .

Step-by-step explanation:

<u>We are given with Outfit 2 , 4 and 6</u>

Outfit 2 - Blue grey red

Outfit 4 - White grey red

Outfit 6 - Black grey red

(A) Since only 2,4, and 6 contain grey slacks, and no other outfit contains grey slacks, those are the subsets that contain grey slacks. Thus , this statement is true .

(B) Just outfits 2, 4, and 6 have grey slacks and a red tie, proving the statement. Other clothes do not include grey slacks and a red tie; they may include a red tie but not both grey slacks and a red tie. Hence , the statement is true .

(C) Since the outfit 2 has grey slacks and a blue shirt, the statement is correct; no other outfit has both grey slacks and a blue shirt. Therefore , this statement is also correct .

(D) The argument is correct since all of the subsets contain grey slacks and no black slacks.

Hence , all the options A , B , C , D are correct and describes the subset .

<u>The given question is incomplete , the complete question is attached here - </u>

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3 years ago

Colin ties 5 groups of balloons to the fence. There are 3 orange balloons

maxonik [38]

Answer:

ok? what is the question?

4 0

3 years ago

Read 2 more answers

A carton of 24 bottles of sports drink costs $16. How many bottles of sports drink can you get

MrRa [10]

Answer:

Step-by-step explanation:

24 bottles = 16 dollars!

? bottles = 12 dollars

Note:

In these type of Questions, find out any dollars you could get with 1 bottle

or in simple terms, (first find the unit Rate)

1 bottle = 1.5 dollars

? bottles = 12 Dollars?

12/1.5

= 8

<h2>I hope that helps! </h2><h3>Have a wonderful day! </h3>

8 0

3 years ago

Find the volume of the rectangular prism.

gregori [183]

Answer:

5 x 3 1/2 x 1 1/2

5 x 7/2 x 3/2 = 26 1/4, in decimal form: 26.25

Multiply across, but first make it all improper fractions.

7 0

3 years ago

You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of

Dima020 [189]

Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

The marble is placed back, and another three red marbles are added

Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

B - R - R

B - R - B

B - B - R

B - B - B

Now, for each case, we have to find the probability that the last marble is red. So

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}$

$P_{1}$ is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is $\frac{30}{50} = \frac{3}{5}$ .

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is $\frac{33}{53}$

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is $\frac{36}{56}$

Again, the red marble sorted is returned, and another 3 are added.

Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is $\frac{39}{59}$ . So

$P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588$

$P_{2}$ is the probability that we go R - R - B - R

$P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788$

$P_{3}$ is the probability that we go R - B - R - R

$P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076$

$P_{4}$ is the probability that we go R - B - B - R

$P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511$

$P_{5}$ is the probability that we go B - R - R - R

$P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733$

$P_{6}$ is the probability that we go B - R - B - R

$P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493$

$P_{7}$ is the probability that we go B - B - R - R

$P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476$

$P_{8}$ is the probability that we go B - B - B - R

$P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419$

So, the probability that this last marble is red is:

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768$

There is a 57.68% probability that this last marble is red.

What's the probability that we actually drew the same marble all four times?

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that we go R-R-R-R. It is the same $P_{1}$ from the previous item(the last marble being red). So $P_{1} = 0.1588$

$P_{2}$ is the probability that we go B-B-B-B. It is almost the same as $P_{8}$ in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.

$P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490$

$P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078$

There is a 20.78% probability that we actually drew the same marble all four times

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3 years ago