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3 years ago

5

A box contains 100 colored chips; some are yellow and some are green. John chooses a chip at random, records the color, and plac

es it back in the bag. John has recorded 57 yellow chips and 19 green chips. Using these results, what is the predicted number of green chips in the box?
25
43
76
81

1 answer:

Andreas93 [3]3 years ago

5 0

Answer:

25

Step-by-step explanation:

A box contains 100 colored chips

Some are yellow and,

Some are green.

John has recorded 57 yellow chips and 19 green chips.

The ratio of yellow chips to green chips is 57:19 simplified becomes; 3:1

The number of green chis is $\frac{1}{3 + 1}$ × 100 = $\frac{1}{4}$ × 100 = 25

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If you have 6 cars, but there is only room in your driveway for 3 cars, in how many ways can you arrange the cars in your drivew

Svetradugi [14.3K]

If the order does not matter, cars can be arranged in 20 ways

If an order is important, cars can be arranged in 120 ways

The probability that the three newest cars end up parked in the driveway is 0.167

1. If the order does not matter, combination is used

$\left ({n} \atop {r}} \right.)=\frac{n!}{r!(n-r)!}$

Here, n=6 r=3

using formula,we get

$\left ( {{6} \atop {3}} \right)=\frac{6!}{3!3!} =5*4=20$

2. If an order is important, Permutation will be applicable

$^{n}P_{r} = \frac{n!}{(n-r)!} \\$

∴ $^{6} P_{3} =\frac{6!}{3!(6-3)!}=\frac{6!}{3!} =120$

3. the probability that the three newest cars end up parked in the driveway

P= $\frac{No. of possible outcomes for 3 cars}{total possibilities}$

= $\frac{6*4*5}{6!}$

= $\frac{120}{720}$

= $\frac{1}{6}$ ≈ 0.167

Hence, If the order does not matter, cars can be arranged in 20 ways

If an order is important, cars can be arranged in 120 ways

The probability that the three newest cars end up parked in the driveway is 0.167

Learn more about probability here brainly.com/question/6077878

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7 0

2 years ago

Help just from 8-20 please please please

sergey [27]

The answer for 13 is 0=12 which means it doesn't have a solution.

3 0

3 years ago

David has 3 candy bars. He wants to cut each candy bars into halves. How many 1/2 sized pieces will he have?

luda_lava [24]

Answer:

6

Step-by-step explanation:

1 candy bar: 2 halves

3 candy bars: 6 halves

Or,

3 ÷ ½ = 6

3 0

4 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

He table shows the price of bagels at two different stores:

galina1969 [7]

Yummy Bagels sells bagels at a lower unit price than Bagel Stop

this is false

5 0

3 years ago

Read 2 more answers