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3 years ago

Katie operates a machine that makes 150 crayons per minute.

Mathematics

2 answers:

Ainat [17]3 years ago

4 0

Answer:

The number of crayons is dependent on the time.

leva [86]3 years ago

3 0

Answer:

The number of crayons is dependent on the time.

Step-by-step explanation:

Alright here is your answer,

The number of crayons is dependent on the time.

Because to know how many crayons you made you would need to know how long that they were made for!

Hope this helps and plz give my brainiest if it's right!

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An online health and medicine company claims that about 50% of Internet users would go to an online sight first for

Serggg [28]

Answer:

The claim that less than 50% of internet users get their health questions answered online is correct, given that according to the survey, 45.98% of people do.

Step-by-step explanation:

Since an online health and medicine company claims that about 50% of Internet users would go to an online sight first for information regarding health and medicine, and a random sample of 1318 Internet users was asked where they will go for information the next time they need information about health or medicine; 606 said that they would use the Internet, to test the claim that less than 50% of Internet users get their health questions answered online, the following calculation must be performed:

1318 = 100

606 = X

606 x 100/1318 = X

60600/1318 = X

45.978 = X

Therefore, the claim that less than 50% of internet users get their health questions answered online is correct, given that according to the survey, 45.98% of people do.

6 0

3 years ago

Priscilla can make 4 bracelets in 17 minutes. At this rate, how many bracelets can she make in 34 minutes?

andrew11 [14]

4 bracelets/17 minutes = x bracelets/34 minutes
cross multiply: 4 · 34 = 17 · x
Divide both sides by 17: (4 · 34)/17 = x
x = 8

Answer: 8 bracelets

6 0

3 years ago

A salesperson earns a commission of $308 for selling $ 2200 in merchandise. Find the commission rate.

Galina-37 [17]

Answer:$247

Step-by-step explanation

5 0

2 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

Pls answer ASAP

ololo11 [35]

The rephrased statement for Kun's proof is: A. In quadrilateral ABCD, if AB ≅ DC & AD ≅ BC, then AB║DC & AD║BC.

<h3>What is a Parallelogram?</h3>

A parallelogram is a quadrilateral that has two opposite sides that are congruent to each other and are also parallel to each other.

This means that if two pairs of opposite sides of a quadrilateral are congruent and parallel, then it is a parallelogram.

Rephrasing Kun's statement in his proof will therefore be: A. In quadrilateral ABCD, if AB ≅ DC & AD ≅ BC, then AB║DC & AD║BC.

Learn more about a parallelogram on:

brainly.com/question/12167853

#SPJ1

8 0

2 years ago