Free 1 Die Roll Calculator - Calculates the probability for the following events in the roll of one fair dice (1 dice roll calculator or 1 die roll calculator):
* Probability of any total from (1-6)
* Probability of the total being less than, less than or equal to, greater than, or greater than or equal to (1-6)
* The total being even
* The total being odd
* The total being a prime number
* The total being a non-prime number
* Rolling a list of numbers i.e. (2,5,6)
* Simulate (n) Monte Carlo die simulations.
1 die calculator

1, 9, 25, 49, .......... What is next 1^2 = 1 3^2 = 9 5^2 = 25 7^2 = 49 So this pattern takes odd numbers and squares them. Our next odd number is 9: 9^2 = [B]81[/B]

2 consecutive odd integers such that their product is 15 more than 3 times their sum. Let the first integer be n. The next odd, consecutive integer is n + 2. We are given the product is 15 more than 3 times their sum: n(n + 2) = 3(n + n + 2) + 15 Simplify each side: n^2 + 2n = 6n + 6 + 15 n^2 + 2n = 6n + 21 Subtract 6n from each side: n^2 - 4n - 21 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-4n-21%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get: n = (-3, 7) If we use -3, then the next consecutive odd integer is -3 + 2 = -1. So we have [B](-3, -1)[/B] If we use 7, then the next consecutive odd integer is 7 + 2 = 9. So we have [B](7, 9)[/B]

Free 2 dice roll Calculator - Calculates the probability for the following events in a pair of fair dice rolls:
* Probability of any sum from (2-12)
* Probability of the sum being less than, less than or equal to, greater than, or greater than or equal to (2-12)
* The sum being even
* The sum being odd
* The sum being a prime number
* The sum being a non-prime number
* Rolling a list of numbers i.e. (2,5,6,12)
* Simulate (n) Monte Carlo two die simulations. 2 dice calculator

3 consecutive odd integers such that thrice the middle is 15 more than the sum of the other 2. [LIST] [*]Let the first integer be n [*]The next odd one (middle) is n + 2. [*]The next odd one is n + 4 [/LIST] We are given 3(n + 2) = n + n + 4 + 15. Simplifying, we get: 3n + 6 = 2n + 19 [URL='http://www.mathcelebrity.com/1unk.php?num=3n%2B6%3D2n%2B19&pl=Solve']Plugging that problem[/URL] into our search engine, we get n = 13. So the next odd integer is 13 + 2 = 15 The next odd integer is 15 + 2 = 17

6 sided die probability to roll a odd number or a number less than 6 First, we'll find the set of rolling an odd number. [URL='https://www.mathcelebrity.com/1dice.php?gl=1&opdice=1&pl=Odds&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']From this dice calculator[/URL], we get: Odd = {1, 3, 5} Next, we'll find the set of rolling less than a 6. [URL='https://www.mathcelebrity.com/1dice.php?gl=4&pl=6&opdice=1&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']From this dice calculator[/URL], we get: Less than a 6 = {1, 2, 3, 4, 5} The question asks for [B]or[/B]. Which means a Union: {1, 3, 5} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5} This probability is [B]5/6[/B]

A is the set of odd integers between 4 and 12 Let A be the set of odd numbers between 4 and 12: [B]A = {5, 7, 9, 11}[/B]

Yes, it was a bit odd. I'll do more research if she comes back on that one. I may have read it wrong. Let me know.

Alex says all factors of 16 are even why is she wrong. [URL='https://www.mathcelebrity.com/factoriz.php?num=16&pl=Show+Factorization']Type in factor 16[/URL] into our search engine. We get the following factor of 16: 1, 2, 4, 8, 16 [B]All of these are even [I]except[/I] 1, which is odd. This is why Alex is wrong.[/B]

[INDENT]Consecutive odd integers are odd integers that differ by ______ , such as ______ and 13. Consecutive even integers are even integers that differ by ______ , such as 12 and ______ . Consecutive odd integers are odd integers that differ by ___2___ , such as ___11___ and 13. Consecutive even integers are even integers that differ by ___2___ , such as 12 and ___14___ .[/INDENT]

2 rules. If either of them passes, then the number is divisible by 11: [LIST=1] [*]Sum of the odd digits - Sum of the even digits is divisible by 11 [*]Sum of the odd digits - Sum of the even digits = 0 (Ex. 121) [/LIST] [MEDIA=youtube]WpV87es0WAU[/MEDIA]

Find four consecutive odd numbers which add to 64. Let the first number be x. The next three numbers are: x + 2 x + 4 x + 6 Add them together to get 64: x + (x + 2) + (x + 4) + (x + 6) = 64 Group like terms: 4x + 12 = 64 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=4x%2B12%3D64&pl=Solve']equation calculator[/URL], we get: [B]x = 13[/B] The next 3 odd numbers are: x + 2 = 13 + 2 = 15 x + 4 = 13 + 4 = 17 x + 6 = 13 + 6 = 19 So the 4 consecutive odd numbers which add to 64 are: [B](13, 15, 17, 19)[/B]

Find the last digit of 4^2081 no calculator 4^1= 4 4^2 = 16 4^3 = 64 4^4 = 256 4^5 = 1024 4^6 = 4096 Notice this pattern alternates between odd exponent powers with the result ending in 4 and even exponent powers with the result ending in 6. Since 2081 is odd, the answer is [B]4. [MEDIA=youtube]ueBWAW4XW4Q[/MEDIA][/B]

Find the odd number less than 100 that is divisible by 9, and when divided by 10 has a remainder of 7. From our [URL='http://www.mathcelebrity.com/divisibility.php?num=120&pl=Divisibility']divisibility calculator[/URL], we see a number is divisible by 9 if the sum of its digits is divisible by 9. Starting from 1 to 99, we find all numbers with a digit sum of 9. This would be digits with 0 and 9, 1 and 8, 2 and 7, 3 and 6, and 4 and 5. 9 18 27 36 45 54 63 72 81 90 Now remove even numbers since the problem asks for odd numbers 9 27 45 63 81 Now, divide each number by 10, and find the remainder 9/10 = 0 [URL='http://www.mathcelebrity.com/modulus.php?num=27mod10&pl=Calculate+Modulus']27/10[/URL] = 2 R 7 We stop here. [B]27[/B] is an odd number, less than 100, with a remainder of 7 when divided by 10.

Find two consecutive odd integers such that the sum of their squares is 290. Let the first odd integer be n. The next odd integer is n + 2 Square them both: n^2 (n + 2)^2 = n^2 + 4n + 4 from our [URL='https://www.mathcelebrity.com/expand.php?term1=%28n%2B2%29%5E2&pl=Expand']expansion calculator[/URL] The sum of the squares equals 290 n^2 + n^2 + 4n + 4 = 290 Group like terms: 2n^2 + 4n + 4 = 290 [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B4n%2B4%3D290&pl=Solve+Quadratic+Equation&hintnum=+0']Enter this quadratic into our search engine[/URL], and we get: n = 11, n = -13 Which means the two consecutive odd integer are: 11 and 11 + 2 = 13. [B](11, 13)[/B] -13 and -13 + 2 = -11 [B](-13, -11)[/B]

Four cousins were born at two-year intervals. The sum of their ages is 36. What are their ages? So the last cousin is n years old. this means consecutive cousins are n + 2 years older than the next. whether their ages are even or odd, we have the sum of 4 consecutive (odd|even) integers equal to 36. We [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof4consecutiveevenintegersis36&pl=Calculate']type this into our search engine[/URL] and we get the ages of: [B]6, 8, 10, 12[/B]

if 2z-1 is an odd integer what is the preceding odd integer? The preceding odd integer is found by subtracting 2: 2z - 1 - 2 [B]2z - 3[/B]

if a and b are odd then a + b is even Let a and b be positive odd integers of the form: [LIST] [*]a = 2n + 1 [*]b = 2m + 1 [/LIST] a + b = 2n + 1 + 2m + 1 a + b = 2n + 2m + 1 + 1 Combing like terms, we get: a + b = 2n + 2m + 2 a + b = 2(n + m) + 2 Let k = n + m a + b = 2k + 2 [B]Therefore a + b is even[/B]

If a is an even integer and b is an odd integer then prove a − b is an odd integer Let a be our even integer Let b be our odd integer We can express a = 2x (Standard form for even numbers) for some integer x We can express b = 2y + 1 (Standard form for odd numbers) for some integer y a - b = 2x - (2y + 1) a - b = 2x - 2y - 1 Factor our a 2 from the first two terms: a - b = 2(x - y) - 1 Since x - y is an integer, 2(x- y) is always even. Subtracting 1 makes this an odd number. [MEDIA=youtube]GDVuQ7bGHx8[/MEDIA]

if m is odd and n is odd, then mn is odd. m = 2k +a where a = 0 or 1 n = 2l + b where b = 0 or 1 mn = (2k + a)(2l + b) = 4kl + 2kb + 2al + ab Since mn is odd, ab = 1 since a = 1 and b = 1

Look at the Contrapositive: If n is even, then 3n + 2 is even... Suppose that the conclusion is false, i.e., that n is even. Then n = 2k for some integer k. Then we have: 3n + 2 = 3(2k) + 2 3n + 2 = 6k + 2 3n + 2 = 2(3k + 1). Thus 3n + 2 is even, because it equals 2j for an integer j = 3k + 1. So 3n + 2 is not odd. We have shown that ¬(n is odd) → ¬(3n + 2 is odd), therefore, the contrapositive (3n + 2 is odd) → (n is odd) is also true.

If n represents an odd integer what represents the previous smaller odd integer Each odd integer is 2 away from the last one, so the previous smaller odd integer is found by subtracting 2 from n: [B]n - 2[/B]

if p=2x is even, then p^2 is also even p^2 = 2 * 2 * x^2 p^2 = 4x^2 This is [B]true [/B]because: [LIST] [*]If x is even, then x^2 is even since two evens multiplied by each other is even and 4x^2 is even [*]If x is odd, the x^2 is odd, but 4 times the odd number is always even since even times odd is even [/LIST]

If x represents the first, or the smaller, of two consecutive odd integers, express the sum of the two integers in terms of x If x is the first of two consecutive odd integers, then we find the next consecutive odd integer by adding 2 to x: x + 2 The sum of the two consecutive odd integers is expressed by x + (x + 2) Simplify by grouping like terms, we get: [B]2x + 2[/B]

Let n be an integer. If n^2 is odd, then n is odd Proof by contraposition: Suppose that n is even. Then we can write n = 2k n^2 = (2k)^2 = 4k^2 = 2(2k) so it is even [I]So an odd number can't be the square of an even number. So if an odd number is a square it must be the square of an odd number.[/I]

Let x be an integer. If x is odd, then x^2 is odd Proof: Let x be an odd number. This means that x = 2n + 1 where n is an integer. [U]Squaring x, we get:[/U] x^2 = (2n + 1)^2 = (2n + 1)(2n + 1) x^2 = 4n^2 + 4n + 1 x^2 = 2(2n^2 + 2n) + 1 2(2n^2 + 2n) is an even number since 2 multiplied by any integer is even So adding 1 is an odd number [MEDIA=youtube]GlzV80M33x0[/MEDIA]

Mr. Jimenez has a pool behind his house that needs to be fenced in. The backyard is an odd quadrilateral shape and the pool encompasses the entire backyard. The four sides are 1818a, 77b, 1111a, and 1919b in length. How much fencing? (the length of the perimeter) would he need to enclose the pool? The perimeter P is found by adding all 4 sides: P = 1818a + 77b + 1111a + 1919b Group the a and b terms P = (1818 + 1111)a + (77 + 1919b) [B]P = 2929a + 1996b[/B]

Free Number Property Calculator - This calculator determines if an integer you entered has any of the following properties:
* Even Numbers or Odd Numbers (Parity Function or even-odd numbers)
* Evil Numbers or Odious Numbers
* Perfect Numbers, Abundant Numbers, or Deficient Numbers
* Triangular Numbers
* Prime Numbers or Composite Numbers
* Automorphic (Curious)
* Undulating Numbers
* Square Numbers
* Cube Numbers
* Palindrome Numbers
* Repunit Numbers
* Apocalyptic Power
* Pentagonal
* Tetrahedral (Pyramidal)
* Narcissistic (Plus Perfect)
* Catalan
* Repunit

n^2+n = odd Factor n^2+n: n(n + 1) We have one of two scenarios: [LIST=1] [*]If n is odd, then n + 1 is even. The product of an odd and even number is an even number [*]If n is even, then n + 1 is odd. The product of an even and odd number is an even number [/LIST]

n^2-n = even Factor n^2-n: n(n - 1) We have one of two scenarios: [LIST=1] [*]If n is odd, then n - 1 is even. The product of an odd and even number is an even number [*]If n is even, then n - 1 is odd. The product of an even and odd number is an even number [/LIST]

Free Odd Numbers Calculator - Shows a set amount of odd numbers and cumulative sum

Free Odds Probability Calculator - Given an odds prediction m:n of an event success, this calculates the probability that the event will occur or not occur

Free Odds Ratio Calculator - This calculator determines the odds ratio for 2 groups X and Y with success and failure for an outcome.

Free Product of Consecutive Numbers Calculator - Finds the product of (n) consecutive integers, even or odd as well. Examples include:
product of 2 consecutive integers
product of 2 consecutive numbers
product of 2 consecutive even integers
product of 2 consecutive odd integers
product of 2 consecutive even numbers
product of 2 consecutive odd numbers
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Use proof by contradiction. Assume sqrt(2) is rational. This means that sqrt(2) = p/q for some integers p and q, with q <>0. We assume p and q are in lowest terms. Square both side and we get: 2 = p^2/q^2 p^2 = 2q^2 This means p^2 must be an even number which means p is also even since the square of an odd number is odd. So we have p = 2k for some integer k. From this, it follows that: 2q^2 = p^2 = (2k)^2 = 4k^2 2q^2 = 4k^2 q^2 = 2k^2 q^2 is also even, therefore q must be even. So both p and q are even. This contradicts are assumption that p and q were in lowest terms. So sqrt(2) [B]cannot be rational. [MEDIA=youtube]tXoo9-8Ewq8[/MEDIA][/B]

Take an integer n. The next alternate consecutive integer is n + 2 Subtract the difference of the squares: (n + 2)^2 - n^2 n^2 + 4n + 4 - n^2 n^2 terms cancel, we get: 4n + 4 Factor out a 4: 4(n + 1) If n is odd, n + 1 is even. 4 * even is always even If n is even, n + 1 is odd. 4 * odd is always odd Since both cases are even, we've proven our statement. [MEDIA=youtube]J_E9lR5qFY0[/MEDIA]

Take an integer n. The next consecutive integer is n + 1 Subtract the difference of the squares: (n + 1)^2 - n^2 n^2 + 2n + 1 - n^2 n^2 terms cancel, we get: 2n + 1 2 is even. For n, if we use an even: we have even * even = Even Add 1 we have Odd 2 is even. For n, if we use an odd: we have even * odd = Even Add 1 we have Odd Since both cases are odd, we've proven our statement. [MEDIA=youtube]RAi0HbH5bqc[/MEDIA]

Take two arbitrary integers, x and y We can express the odd integer x as 2a + 1 for some integer a We can express the odd integer y as 2b + 1 for some integer b x + y = 2a + 1 + 2b + 1 x + y = 2a + 2b + 2 Factor out a 2: x + y = 2(a + b + 1) Since 2 times any integer even or odd is always even, then [B]x + y by definition is even[/B]. [MEDIA=youtube]9A-qe4yZXYw[/MEDIA]

Let us take an integer x which is both even [I]and[/I] odd. [LIST] [*]As an even integer, we write x in the form 2m for some integer m [*]As an odd integer, we write x in the form 2n + 1 for some integer n [/LIST] Since both the even and odd integers are the same number, we set them equal to each other 2m = 2n + 1 Subtract 2n from each side: 2m - 2n = 1 Factor out a 2 on the left side: 2(m - n) = 1 By definition of divisibility, this means that 2 divides 1. But we know that the only two numbers which divide 1 are 1 and -1. Therefore, our original assumption that x was both even and odd must be false. [MEDIA=youtube]SMM9ubEVcLE[/MEDIA]

Roberto has taken 17 photos photos are placed on each odd number page and the newspaper has 10 pages total. The pages with photographs will have 3 or 4 photos each. How many pages has 3 photos and how many pages have 4 photos? Odd pages are 1, 3, 5, 7, 9 17/5 = 3 with 2 remaining. So all 5 pages have 3 photos. Then with 2 left over, 2 pages get 4 photos. So 5 pages have [B]3 photos, and 2 pages have 2 photos[/B] 3(3) + 4(2) = 9 + 8 = 17

Sam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783. Which equation could be used to find Jeremy's age, j, if he is the younger man. Let Sam's age be s. Let' Jeremy's age be j. We're given: [LIST=1] [*]s = j + 2 <-- consecutive odd integers [*]sj = 783 [/LIST] Substitute (1) into (2): (j + 2)j = 783 j^2 + 2j = 783 Subtract 783 from each side: j^2 + 2j - 783 = 0 <-- This is the equation to find Jeremy's age. To solve this, [URL='https://www.mathcelebrity.com/quadratic.php?num=j%5E2%2B2j-783%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type this quadratic equation into the search engine[/URL] and get: j = 27, j = -29. Since ages cannot be negative, we have: [B]j = 27[/B]

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sum of 3 consecutive odd integers equals 1 hundred 17 The sum of 3 consecutive odd numbers equals 117. What are the 3 odd numbers? 1) Set up an equation where our [I]odd numbers[/I] are n, n + 2, n + 4 2) We increment by 2 for each number since we have [I]odd numbers[/I]. 3) We set this sum of consecutive [I]odd numbers[/I] equal to 117 n + (n + 2) + (n + 4) = 117 [SIZE=5][B]Simplify this equation by grouping variables and constants together:[/B][/SIZE] (n + n + n) + 2 + 4 = 117 3n + 6 = 117 [SIZE=5][B]Subtract 6 from each side to isolate 3n:[/B][/SIZE] 3n + 6 - 6 = 117 - 6 [SIZE=5][B]Cancel the 6 on the left side and we get:[/B][/SIZE] 3n + [S]6[/S] - [S]6[/S] = 117 - 6 3n = 111 [SIZE=5][B]Divide each side of the equation by 3 to isolate n:[/B][/SIZE] 3n/3 = 111/3 [SIZE=5][B]Cancel the 3 on the left side:[/B][/SIZE] [S]3[/S]n/[S]3 [/S]= 111/3 n = 37 Call this n1, so we find our other 2 numbers n2 = n1 + 2 n2 = 37 + 2 n2 = 39 n3 = n2 + 2 n3 = 39 + 2 n3 = 41 [SIZE=5][B]List out the 3 consecutive odd numbers[/B][/SIZE] ([B]37, 39, 41[/B]) 37 ← 1st number, or the Smallest, Minimum, Least Value 39 ← 2nd number 41 ← 3rd or the Largest, Maximum, Highest Value

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Free Sum of the First (n) Numbers Calculator - Determines the sum of the first (n)
* Whole Numbers
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* Even Numbers
* Odd Numbers
* Square Numbers
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Sum of two consecutive numbers is always odd Definition: [LIST] [*]A number which can be written in the form of 2 m where m is an integer, is called an even integer. [*]A number which can be written in the form of 2 m + 1 where m is an integer, is called an odd integer. [/LIST] Take two consecutive integers, one even, and one odd: 2n and 2n + 1 Now add them 2n + (2n+ 1) = 4n + 1 = 2(2 n) + 1 The sum is of the form 2n + 1 (2n is an integer because the product of two integers is an integer) Therefore, the sum of two consecutive integers is an odd number.

Take a look at the following sums: 1 = 1 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 a. Come up with a conjecture about the sum when you add the first n odd numbers. For example, when you added the first 5 odd numbers (1 + 3 + 5 + 7 + 9), what did you get? What if wanted to add the first 10 odd numbers? Or 100? b. Can you think of a geometric interpretation of this pattern? If you start with one square and add on three more, what can you make? If you now have 4 squares and add on 5 more, what can you make? c. Is there a similar pattern for adding the first n even numbers? 2 = 2 2 + 4 = 6 2 + 4 + 6 = 12 2 + 4 + 6 + 8 = 20 a. The formula is [B]n^2[/B]. The sum of the first 10 odd numbers is [B]100[/B] seen on our s[URL='http://www.mathcelebrity.com/sumofthefirst.php?num=10&pl=Odd+Numbers']um of the first calculator[/URL] The sum of the first 100 odd numbers is [B]10,000[/B] seen on our [URL='http://www.mathcelebrity.com/sumofthefirst.php?num=100&pl=Odd+Numbers']sum of the first calculator[/URL] b. Geometric is 1, 4, 9 which is our [B]n^2[/B] c. The sum of the first n even numbers is denoted as [B]n(n + 1)[/B] seen here for the [URL='http://www.mathcelebrity.com/sumofthefirst.php?num=+10&pl=Even+Numbers']first 10 numbers[/URL]

The set of all odd numbers between 10 and 30 [B]{11, 13, 15, 17, 19, 21, 23, 25, 27, 29}[/B]

The sum of 5 odd consecutive numbers is 145. Let the first odd number be n. We have the other 4 odd numbers denoted as: [LIST] [*]n + 2 [*]n + 4 [*]n + 6 [*]n + 8 [/LIST] Add them all together n + (n + 2) + (n + 4) + (n + 6) + (n + 8) The sum of the 5 odd consecutive numbers equals 145 n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 145 Combine like terms: 5n + 20 = 145 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=5n%2B20%3D145&pl=Solve']equation solver[/URL], we get [B]n = 25[/B]. Using our other 4 consecutive odd numbers above, we get: [LIST] [*]27 [*]29 [*]31 [*]33 [/LIST] Adding the sum up, we get: 25 + 27 + 29 + 31 + 33 = 145. So our 5 odd consecutive number added to get 145 are [B]{25, 27, 29, 31, 33}[/B]. [MEDIA=youtube]0T2PDuQIIwI[/MEDIA]

There are 15 cards, numbered 1 through 15. If you pick a card, what is the probability that you choose an odd number or a two? We want the P(odd) or P(2). P(odd) = 1, 3, 5, 7, 9, 11, 13, 15 = 8/15 P(2) = 1/15 Add them both: 8/15 + 1/15 = 9/15 Simplified, we get [B]3/5[/B].

There is a stack of 10 cards, each given a different number from 1 to 10. Suppose we select a card randomly from the stack, replace it, and then randomly select another card. What is the probability that the first card is an odd number and the second card is greater than 7. First Event: P(1, 3, 5, 7, 9) = 5/10 = 1/2 or 0.5 Second Event: P(8, 9, 10) = 3/10 or 0.3 Probability of both events since each is independent is 1/2 * 3/10 = 3/20 = [B]0.15 or 15%[/B]

Three good friends are in the same algebra class, their scores on a recent test are three consecutive odd integers whose sum is 273. Find the score In our search engine, we type in [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=3consecutiveintegerswhosesumis273&pl=Calculate']3 consecutive integers whose sum is 273[/URL] and we get: [B]90, 91, 92[/B]

Todd bought 5 ice cream sandwiches for $3.75. Bryce bought one ice cream sandwich for $1.00. Who got the better deal? Todd's unit cost is found by: Todd's Unit Cost = Total Price / Total Ice Cream Sandwiches Todd's Unit Cost = $3.75/5 Todd's Unit Cost = $0.75 Bryce's unit cost is $1.00 per ice cream sandwich, so [B]Todd got the better deal.[/B]

todd has a bag of 150 pieces of candy. every weekday he eats 2 pieces and 3 pieces on weekends until he has no more left. So each week, he eats 2*5 + 3*2 = 10 + 6 = 16 pieces 16 per week * 9 weeks = 144 pieces 150 - 144 = 6 pieces left 3 weekdays * 2 pieces per weekday = 6. So, Todd ate all the candy in 9 weeks, 3 days.

Two dice are rolled. Enter the size of the set that corresponds to the event that both dice are odd. If dice 1 is odd, then we have the following face values: {1, 3, 5} If dice 2 is odd, then we have the following face values: {1, 3, 5} [URL='https://www.mathcelebrity.com/2dice.php?gl=1&opdice=1&pl=Both+Odd&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']From this 2 dice odds face link[/URL], we see that the size of the set is 9. [LIST=1] [*]{1, 1} [*]{1, 3} [*]{1, 5} [*]{3, 1} [*]{3, 3} [*]{3, 5} [*]{5, 1} [*]{5, 3} [*]{5, 5} [/LIST]

You are using a spinner with the numbers 1-10 on it. Find the probability that the pointer will stop on an odd number or a number less than 4. We want P(odd number) or P(n<4). [LIST] [*]Odd numbers are {1, 3, 5, 7, 9} [*]n < 4 is {1, 2, 3} [/LIST] We want the union of these 2 sets: {1, 2, 3, 5, 7, 9} We have 6 possible pointers in a set of 10. [B]6/10 = 3/5 = 0.6 or 60%[/B]

P(X = 4) AND P(X is odd) Which means P(X = 1) + P(X = 3) [LIST] [*]P(X = 1) = 1/6 [*]P(X = 3) = 1/6 [*]P(X = 1) + P(X = 3) = 1/6 + 1/6 = 2/6 = [B]1/3[/B] [/LIST]

You roll a red die and a green die. What is the size of the sample space of all possible outcomes of rolling these two dice, given that the red die shows an even number and the green die shows an odd number greater than 1? [LIST] [*]Red Die Sample Space {2, 4, 6} [*]Green Die Sample Space {3, 5} [*]Total Sample Space {(2, 3), (2, 5), (4, 3), (4, 5), (6, 3), (6, 5)} [*]The sie of this is 6 elements. [/LIST]