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3 years ago

The cup dispenser at a water cooler contains cone-shaped paper cups with the

Mathematics

1 answer:

Crank3 years ago

6 0

48 cubic centimeters

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i kind of know its C need 2nd opinion A tire manufacturer took a random sample of 50 tires and found that 3 were defective. In a

den301095 [7]

It's kind of not C.

Solve this proportion for 'x':

3/50 = x/2400

Hint: Cross-multiply the proportion, then do what you need to do.

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3 years ago

A bag holds 12 red marbles, 11 green

Blizzard [7]

Answer:

62.22%

Step-by-step explanation:

First, find how many total marbles there are:

12 + 11 + 17 + 5

= 45

Then, find how many marbles that are not blue:

12 + 11 + 5

= 28

Then, divide 28 by 45, and multiply by 100

(28/45) x 100

= 62.22% is the probability of not choosing a blue marble

3 0

3 years ago

Read 2 more answers

AB and DE are chords that intersect at point F inside circle C as shown. If the measure of arc EA= 50 degrees and the measure of

defon

Answer:

Part 1) $m\angle AFE=55^o$

Part 2) $m\angle EFB=125^o$

Step-by-step explanation:

Part 1) what is the measure of angle AFE

we know that

The measure of the interior angle is the semisum of the arches that comprise it and its opposite.

<u>Note:</u> In this problem the correct measure of arc EA is 40 degrees (see the picture)

$m\angle AFE=\frac{1}{2}(arc\ EA+arc\ DB)$

substitute the given values

$m\angle AFE=\frac{1}{2}(40^o+70^o)=55^o$

Part 2) what is the measure of angle EFB?

we know that

$m\angle AFE+m\angle EFB=180^o$ ---> by supplementary angles (form a linear pair)

substitute the given value

$55^o+m\angle EFB=180^o$

$m\angle EFB=180^o-55^o=125^o$

8 0

4 years ago

Kurt johnson delivers packages for a delivery company. he receives $7.50 for every package delivered. what does kurt earn in a w

Ivan

Y = 7.50x.....y = earnings per week , x = number of packages delivered

48 packages.....so x = 48
y = 7.50(48)
y = 360.....so if he delivers 48 packages, he earns $ 360

8 0

3 years ago

Read 2 more answers

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