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algol13
2 years ago
12

Mr. Roper has eight teams of students in his debate class. Their grades on the last debate assignment are denoted by the set (99

, 72,
86, 80, 96, 87, 79, 87). Which 5-number summary represents the data set?
{72, 79, 86, 96, 99}
{72, 79.5, 86.5, 91.5, 99}
{72, 79, 83, 87, 96)
{72, 79, 86, 87, 96}
Mathematics
1 answer:
muminat2 years ago
4 0

Answer:

l think 3rd option must be answer

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Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145

5 0
2 years ago
Henry needs 2 pints of red paint and 3 pints of yellow paint to get a specific shade of orange. If he uses 9 pints of yellow pai
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Tickets to a play are $12.00 for adults. Children receive a discount and only have to pay $8.00. If 40 people attend the play an
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30 were adults and 10 were children :) just add it up
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A certain organization recommends the use of passwords with the following​ format: vowel commavowel, consonant comma consonant c
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The password format is = inpray93

vowel , consonant, consonant, consonant, vowel, consonant, number, number.

There are 21 consonants, 5 vowels and 10 number choices (0-9)

Since, there are 21 consonants and repetition is allowed so for 2nd, 3rd, 4th, and 6th positions 21 choices are available. For 1st and 5th positions, 5 choices  are available and for 7th and 8th position there are 10 choices (0 to 9).

Passwords are not case sensitive which means upper case and lower case letters are same.

So, the number of passwords can be =

5\times21\times21\times21\times5\times21\times10\times10

= 486202500

Part (b) question is not given. But it can be opposite to the first part i.e when the password letters are case sensitive.

This means all the letters are different.

So, number of passwords will be =

10\times42\times42\times42\times10\times42\times10\times10

= 31116960000

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2 years ago
zoe and Hannah share tips in the ratio 3:7 last week zoe recived £24 how much did hannah recieve last week
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Answer:

56

Step-by-step explanation:

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