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FrozenT [24]
3 years ago
8

There are 12 tables and 4 booths in the family restaurant. Each table seats 4 people. If the restaurant can seat up to 72 people

, what is the capacity of each booth?
5 people


6 people


4 people


8 people
Mathematics
2 answers:
Bezzdna [24]3 years ago
5 0

Answer:

6

Step-by-step explanation:

12 times 4 = 48

72 - 48 = 24

24 divided by 4 = 6

kupik [55]3 years ago
5 0

Answer:

4

Step-by-step explanation:

You might be interested in
Mrs. Clements has a table with a top that is shaped like the trapezoid shown. Which expression can Mrs. Clements use to find the
Lunna [17]

Answer:

60

Step-by-step explanation:

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

3 0
3 years ago
Someone please help me
Maksim231197 [3]

Answer:

c

Step-by-step explanation:

5 0
2 years ago
A rectangular painting is 24 inches wide and 20 inches tall without the frame. With the frame, it is 28 inches wide and 24 inche
evablogger [386]

The area of the frame not covered by the painting is 192 square inches

<em><u>Solution:</u></em>

Given that, A rectangular painting is 24 inches wide and 20 inches tall without the frame

The area of rectangle is given as:

Area = length \times width

<em><u>Find the area of painting without frame:</u></em>

Length = 20 inches

Width = 24 inches

Area\ Without\ Frame = 20 \times 24 = 480

Thus area of painting without frame is 480 square inches

With the frame, it is 28 inches wide and 24 inches tall

<em><u>Find the area of painting with frame:</u></em>

Length = 24 inches

Width = 28 inches

Area\ with\ frame = 24 \times 28 = 672

Thus area of painting with frame is 672 square inches

<em><u>What is the area of the frame not covered by the painting</u></em>

Area of the frame not covered by the painting = Area with farme - Area without frame

Area of the frame not covered by the painting = 672 - 480 = 192

Thus the area of the frame not covered by the painting is 192 square inches

4 0
3 years ago
30 points.Please answer with explanation​
Bas_tet [7]

Answer:

m∠1 = 142

Step-by-step explanation:

38 + m∠1 = 180

m∠1 = 142

5 0
3 years ago
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