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maksim [4K]
3 years ago
8

Middle School: Math (10 Points)

Mathematics
1 answer:
sasho [114]3 years ago
5 0
1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:

A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?

You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked

2) a - 2%  as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.  

b - 1,000,000/2500 = 400
400 packages are defective each year
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A rich woman deposits a whole number of dollars x in her bank. The next time she deposits y dollars (still a whole number). Each
grin007 [14]

Answer:

x = 12

y = 22

Total amount = $2596

Step-by-step explanation:

First let's find the value of each deposit until the 10th in relation to x and y:

1st: x

2nd: y

3rd: x + y

4th: x + 2y

5th: 2x + 3y

6th: 3x + 5y

7th: 5x + 8y

8th: 8x + 13y

9th: 13x + 21y

10th: 21x + 34y

Now, we can write a system with two equations and two variables:

2x + 3y = 90

21x + 34y = 1000

From the first equation: x = (90 - 3y)/2

Using this value of x in the second equation, we have:

21*(90 - 3y)/2 + 34y = 1000

945 - 31.5y + 34y = 1000

2.5y = 55

y = 22

Now we can find x:

x = (90 - 3*22)/2 = 12

Now, summing all the deposits, we have a total of 55x + 88y, which is equal to 55*12 + 88*22 = $2596

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3 years ago
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yulyashka [42]
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Step-by-step explanation:

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Most ATM’s require a 4-digit code using the digits 0-9 how many 4-digit codes can you use
Zolol [24]

10000 digits can be used for 4 digit A.T.M code.

<u>Solution:</u>

Given that A.T.M required 4 digit codes using the digits 0 to 9.  

Need to determine how many four digit code can be used.

We are assuming that number starting with 0 are also valid ATM codes that means 0789 , 0089 , 0006 and 0000 are also valid A.T.M codes.

Now we have four places to be filled by 0 to 9 that is 10 numbers

Also need to keep in mind that repetition is allowed in this case means if 9 is selected at thousands place than also it is available for hundreds, ones or tens place .  

First digit can be selected in 10 ways that is from 0 to 9.

After selecting first digit, second digit can be selected in 10 ways that is 0 to 9 and same holds true for third and fourth digit.

So number of ways in which four digit number is created = 10 x 10 x 10 x 10 = 10000 ways

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