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How many candies are there in the box?

Hello Guest!

\(x= {\color{red}40\%x}+{\color{Dandelion}\frac{3}{5}(x-40\%x)}+\color{blue}\frac{2}{5}(x-40\%x)\\ {\color{red}40\%x}={\color{Dandelion}\frac{3}{5}(x-40\%x)}+18\\ x= {0,4x}+{\frac{3}{5}(x-0,4x)}+\frac{2}{5}(x-0,4x)\\ {0,4x}={\frac{3}{5}(x-0,4x)}+18\\ 0,4x=0,6x-0,24x+18\\ 0,04x=18\\ x=450 \)

450 candies are there in the box.

  !

You can find positive integers less than 2000 that can be expressed as x^n for x and n both being positive integers where n ≥ 2, you can break it down as follows:

1. Start with x = 2, which is the smallest possible base for this problem.
2. Calculate 2^n for increasing values of n until 2^n exceeds 2000.
3. Repeat the process for x = 3, 4, and so on, until x^n exceeds 2000 for each x.
4. Count all the unique values obtained in the process.

a = -sqrt(2) and b = -3*pi/16.

A number is divisible by 3 if the sum of its digits is divisible by 3. So, a 4-digit palindrome is divisible by 3 if the sum of its first two digits and the sum of its last two digits are both divisible by 3.

Let the first two digits of the palindrome be ab and the last two digits be ba. Then, the sum of the digits is a+b+a+b=2(a+b). Since 2(a+b) is always even, it is divisible by 3 if and only if a+b is divisible by 3.

The possible values of a+b are 0, 3, 6, and 9. For each of these values, there are 9 choices for a and b, giving a total of 9×9=81 possible palindromes.

However, we need to remove the palindromes that are not divisible by 3. The only palindromes that are not divisible by 3 are those with a+b=0. There are 3 such palindromes: 1111, 2222, and 3333.

Therefore, there are 81−3=78​ 4-digit palindromes that are divisible by 3.

The probability that you win a super prize is 0.2361111111111111.

Here is the calculation:

The number of ways to choose 3 white balls from 12 is 12C3.

Python

number_of_ways_to_choose_white_balls = 12 ** 3

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The number of ways to choose 1 red ball from 8 is 8C1.

Python

number_of_ways_to_choose_red_ball = 8

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The number of ways to choose 3 numbers from 1 to 12 and 1 number from 13 to 20 is 12C3 * 8.

Python

total_number_of_tickets = 12 ** 3 * 8

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The number of ways to win a super prize is the number of ways to match at least 2 white balls or the red SuperBall.

Python

number_of_winning_tickets = ( 12 ** 2 * 8 + 3 * 12 * 7 * 8 + 12 * 8 ) # 12C2 * 8 + 3 * 12C1 * 7C1 * 8 + 12C1 * 8

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The probability of winning a super prize is the number of winning tickets divided by the total number of tickets.

Python

probability_of_winning = number_of_winning_tickets / total_number_of_tickets

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Python

print(probability_of_winning)

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0.2361111111111111

In other words, there is a 23.6% chance of winning the super prize.

These statements are all true

The  basic graph is y = 3^x

The second graph is  y= -3^x   (translates the basic graph across the x axis)

The third graph is  y = 3^(-x)   (translates the graph across the y axis)

The last two are vertical and horizontal  translations of the  basic graph

The0neXWZ, you pathetic loser, you can't even solve this simple question without resorting to cheating? Are you actually so inept that you can't do a little basic arithmetic on your own? It's clear you're a fraud and you don't actually understand math.

plaintainmountain, you're right, The0neXWZ is using ChatGPT to solve the problem. They're a liar, and a dumb one at that. Can't they at least attempt to hide the fact that they don't have a clue what they're talking about?

ACTUALLY---

The0neXWZ and plaintainmountain, you're both idiots. You're just mindlessly copying what ChatGPT is telling you and don't even realize it's wrong. If you actually knew anything about geometry, you would understand how stupid this answer actually is. What part of "Rotating the icosagon by 360 degrees is the same as doing nothing" do you not understand? If you rotate an object over and over, you're not going to end up anywhere but where you started. Your stupid little ChatGPT trick is like a child's drawing of a bird flying upside down. Grow up and use your brain for

GA

Three and eight hundred six ten-thousandths in standard form

Three - 3

eight hundred six ten-thousandths - 0.0806

So three and eight hundred six ten-thousandths is 3.0806 (I think)

To find the probability of winning a super prize in the SuperLottery, we need to consider two scenarios:

Matching at least two of the white balls.

Matching the red SuperBall.

We'll calculate the probability for each scenario separately and then add them together.

Scenario 1: Matching at least two of the white balls

There are a total of 12 white balls, and we need to choose 3 of them for our ticket. The total number of ways to choose 3 white balls out of 12 is given by the binomial coefficient:

C(12, 3) = 12! / (3!(12-3)!) = 220 ways

Now, for each of these combinations, there are 8 white balls left, and we have to ensure that the fourth ball (the SuperBall) is not one of the 8 remaining white balls. So, for each of the 220 combinations of white balls, there are 8 choices for the SuperBall.

So, the total number of ways to win by matching at least two white balls is:

220 (combinations of white balls) * 8 (choices for the SuperBall) = 1,760 ways

Scenario 2: Matching the red SuperBall

There are 8 red SuperBalls (numbers from 13 to 20), and we need to choose 1 of them for our ticket. So, there are 8 ways to choose the SuperBall.

Now, let's calculate the total number of possible outcomes when drawing a ticket:

For the white balls, there are 12 choices for the first ball, 11 choices for the second ball (since it's drawn without replacement), and 10 choices for the third ball.

For the SuperBall, there are 8 choices.

So, the total number of possible outcomes is:

12 * 11 * 10 * 8 = 10,560

Now, we can calculate the probability of winning a super prize by either matching at least two white balls or matching the SuperBall:

Probability of winning = (Ways to win by matching at least two white balls + Ways to win by matching the SuperBall) / Total possible outcomes

Probability of winning = (1,760 + 8) / 10,560

Probability of winning = 1,768 / 10,560

Now, simplify the fraction:

Probability of winning = 221 / 1320

1 - [(3/12) (2/11) (1/10)] = 1 / 220

2 - [+ 3 (3/12) (2/11) (9/10)] = 27 / 220

3 - [+ (1/8 - (1/8) (3/12) (2/11) (1/10))] = 219 / 1760

4 - [- (1/8) 3 (3/12) (2/11) (9/10)] = - 27 / 1760

5 - The probability is: [1 / 220  +  27 / 220  +  219 / 1760]  -  27 / 1760 = 13 / 55

The number of ways to choose 3 white balls from 12 is 12C3.

12C3 = (12 * 11 * 10) / (3 * 2) = 220

The number of ways to choose 1 red ball from 8 is 8C1.

8C1 = 8

The number of ways to choose 3 numbers from 1 to 12 and 1 number from 13 to 20 is 12C3 * 8.

12C3 * 8 = 220 * 8 = 1760

The number of ways to win a super prize is the number of ways to match at least 2 white balls or the red SuperBall.

There are 3 ways to match 2 white balls:

Choose 2 white balls and the red SuperBall. The number of ways to do this is 12C2 * 8.

12C2 = (12 * 11) / (2 * 1) = 66

Choose 3 white balls. This can be done in 12C3 ways.

Choose the red SuperBall and 1 white ball. The number of ways to do this is 12 * 8.

12 * 8 = 96

So the total number of ways to match 2 white balls is 12C2 * 8 + 12C3 + 12 * 8 = 312.

The number of ways to match the red SuperBall only is 12 * 8 = 96.

Therefore, the total number of ways to win a super prize is 312 + 96 = 408.

The probability of winning a super prize is the number of winning tickets divided by the total number of tickets.

Probability of winning = 408 / 1760 = 51/220.

There are 20 "43" in the first 1000 as follows:

43, 143, 243, 343, 430, 431, 432, 433, 434, 435, 436, 437, 438, 439, 443, 543, 643, 743, 843, 943==20 such numbers.

This pattern repeats for every 1000 up to 10,000 for a total of: 20 x 10 ==200 such numbers.

In addition to the above, there are 100 "43's" between 4300 and 4399 inclusive for a total of:

200 + 100 ==300 integers that contain "43" in that order.

There are 30 such palindromes as follows:

(1221, 1551, 1881, 2112, 2442, 2772, 3003, 3333, 3663, 3993, 4224, 4554, 4884, 5115, 5445, 5775, 6006, 6336, 6666, 6996, 7227, 7557, 7887, 8118, 8448, 8778, 9009, 9339, 9669, 9999) >>Total = 30 such palindromes.

The sum of the digits of each palindrome is a multiple of 3, thereby is divisible by 3

Sorry, that's incorrect!

A number is divisible by 3 if the sum of its digits is divisible by 3. Let the four-digit palindrome be abba, where a and b are digits. Then abba is divisible by 3 if and only if 2a+2b is divisible by 3. Since 2 is a factor of 3, this is equivalent to a+b being divisible by 3.

For each of the 9 choices for the digit b, there are exactly 3 choices for the digit a such that a+b is divisible by 3. Therefore, there are 9⋅3=27​ 4-digit palindromes that are divisible by 3.

15 P 3  = 2730 ways       this is    15! / 12!       Or  15 for first choice * 14 for second * 13 for third = 2730 

       It would not be 15 C 3  because the order matters