24 coloring books, 60 crayons, and 84 markers can be packaged into at most how many identical packages? How many of each would each package contain? First, determine the greatest common factor (GCF) of 24, 60, and 84 using our [URL='http://www.mathcelebrity.com/gcflcm.php?num1=24&num2=60&num3=84&pl=GCF']GCF calculator[/URL]. GCF(24, 60, 84) = 12 So we have 12 identical packages. Now, figure out how many coloring books, crayons, and markers for each package [LIST] [*]24/12 = 2 coloring books [*]60/12 = 5 crayons [*]84/12 = 7 markers [/LIST] [B]So we have 12 identical packages, each containing 2 coloring books, 5 crayons, and 7 markers[/B]

A bag contains 10 red balls, 10 green balls and 6 white balls. Two balls are drawn at random from the bag without replacement. What is the probability that they are of different colours? [LIST] [*]The key phrase here is [I]without replacement[/I]. [*]First, it's easier to find the probability of both colors matching, and then subtracting that from 1. [/LIST] We want 1 - (P(Red-Red) + P(Green-Green) + P(White-White)). So we have the following: [U]Find the probability of both colors matching[/U] P(Red-Red) = 10/26 * 9/25 = 0.138462 P(Green-Green) = 10/26 * 9/25 = 0.138462 P(White-White) = 6/26 * 5/25 = 0.046154 P(Red-Red) + P(Green-Green) + P(White-White) = 0.13846 + 0.13846 + 0.046154 = 0.3231 Now, we want to take the complement of this probability which is no colors matching, so we have: P(Both Different Colors) = 1 - P(Same Colors) P(Both Different Colors) = 1 - 0.3231 P(Both Different Colors) = [B]0.6769[/B]

A bag contains 3 red marbles and 4 blue marbles. a marble is taken at random and replaced. Another marble is taken from the bag. Work out the probability that the two marbles taken from the bag are the same color. [LIST] [*]Total number of marbles in the bag is 3 + 4 = 7. [*]The problem asks for the probability of (RR) [I]or[/I] (BB). [*]It's worthy to note we are replacing the balls after each draw, which means we always have 7 to draw from [/LIST] Since each draw is independent, we take the product of each event for the total event probability. P(RR) = 3/7 * 3/7 = 9/49 P(BB) = 4/7 * 4/7 = 16/49 We want to know P(RR) + P(BB) P(RR) + P(BB) = 9/49 + 16/49 = 25/49 [MEDIA=youtube]26F9vjsgNGs[/MEDIA]

A bag contains 8 marbles of different colors. How many ways unique ways or orders can you select 3 marbles? We want the combinations formula, 8 choose 3, or 8C3. [URL='https://www.mathcelebrity.com/permutation.php?num=8&den=3&pl=Combinations']Type 8 C 3 into our search engine and we get:[/URL] [B]56 unique ways[/B]

A belongs to Both X and Y What this means is Element A belongs to Set X and Set Y. We write this as follows A is an element of X is written as A ? X A is an element of Y is written as A ? Y [B]A ? X & A ? Y[/B] [FONT=Droid Serif][COLOR=rgb(34, 34, 34)][SIZE=14px][/SIZE][/COLOR][/FONT]

A box contains 4 plain pencils and 4 pens. A second box contains 5 color pencils and 3 crayons. One item from each box is chosen at random. What is the probability that a pen from the first box and a crayon from the second box are selected? [LIST] [*]First box, P(pen) = 4/8 = 1/2 = 0.5 [*]Second box, P(crayon) = 3/8 [/LIST] Since each event is independent, we have: P(Pen from Box 1) * P(Crayon from Box 2) = 1/2 * 3/8 = [B]3/16 or 0.1875[/B]

A box contains 10 bells. There are 6 red bells and the rest are silver. What is the probability of picking two bells of the same color if the bell is replaced after each pick? If there are 6 red bells, then we have 10 - 6 = 4 silver bells. The problem asks for the probability of picking two bells of the same color. Which mean we have 2 scenarios: [LIST=1] [*]Silver, Silver [*]Red, Red [/LIST] Find the probability of Silver, Silver: Since each draw is independent, and we replace the bells, we have a 4/10 chance of picking silver. [URL='https://www.mathcelebrity.com/fraction.php?frac1=4%2F10&frac2=3%2F8&pl=Simplify']Simplified, this is 2/5[/URL]. (2/5)(2/5) = 4/25 Find the probability of Red, Red: Since each draw is independent, and we replace the bells, we have a 6/10 chance of picking silver. [URL='https://www.mathcelebrity.com/fraction.php?frac1=6%2F10&frac2=3%2F8&pl=Simplify']Simplified, this is 3/5[/URL]. (3/5)(3/5) = 9/25 Because we want Silver, Silver [B][U]or[/U][/B] Red, Red, we add the two probabilities. 4/25 + 9/25 = [B]13/25[/B]

A box contains 4 plain pencils and 4 pens. A second box contains 5 color pencils and 3 crayons. One item from each box is chosen at random. What is the probability that a pen from the first box and a crayon from the second box are selected? [LIST] [*]First box, P(pen) = 4/8 = 1/2 = 0.5 [*]Second box, P(crayon) = 3/8 [/LIST] Since each event is independent, we have: P(Pen from Box 1) * P(Crayon from Box 2) = 1/2 * 3/8 = [B]3/16 or 0.1875[/B]

A box contains 5 plain pencils and 3 pens. A second box contains 2 color pencils and 2 crayons. One item from each box is chosen at random. What is the probability that a plain pencil from the first box and a color pencil from the second box are selected [U]Calculate the probability of a plain pencil in the first box:[/U] P(plain pencil in the first box) = Total Pencils / Total Objects P(plain pencil in the first box) = 5 pencils / (5 pencils + 3 pens) P(plain pencil in the first box) = 5/8 [U]Calculate the probability of a color pencil in the first box:[/U] P(color in the second box) = Total Pencils / Total Objects P(color in the second box) = 2 pencils / (2 pencils + 2 crayons) P(color in the second box) = 2/4 We can simplify this. [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F4&frac2=3%2F8&pl=Simplify']Type 2/4 into our search engine[/URL] and we get 1/2 Now the problem asks for the probability that a plain pencil from the first box and a color pencil from the second box are selected. Since each event is independent, we multiply them together to get our answer: P(plain pencil in the first box, color in the second box) = P(plain pencil in the first box) * P(color in the second box) P(plain pencil in the first box, color in the second box) = 5/8 * 1/2 P(plain pencil in the first box, color in the second box) = [B]5/16[/B]

A box contains 5 plain pencils and 7 pens. A second box contains 4 color pencils and 4 crayons. One item from each box is chosen at random. What is the probability that a plain pencil from the first box and a color pencil from the second box are selected? Probability of plain pencil from first box: 5/(5 + 7) = 5/12 Probability of color pencil from second box: 4/(4 + 4) = 4/8 = 1/2 Probability of both events together: Since each event is independent, we multiply probabilities: 5/12 * 1/2 = [B]5/24[/B]

A box contains 6 yellow, 3 red, 5 green, and 7 blue colored pencils. A pencil is chosen at random, it is not replaced, then another is chosen. What is the probability of choosing a red followed by a green? We have 6 + 3 + 5 + 7 = 21 total pencils P(Red on the first draw) = Total Red / Total pencils P(Red on the first draw) = 3/21 [URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F21&frac2=3%2F8&pl=Simplify']P(Red on the first draw)[/URL] = 1/7 We're drawing without replacement, this means on the next draw, we have 21 - 1 = 20 pencils P(Green on the second draw) = Total Green / Total pencils P(Green on the second draw) = 5/20 [URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F20&frac2=3%2F8&pl=Simplify']P(Green on the second draw) [/URL]= 1/4 Since each event is independent, we have: P(Red on first, green on second) = P(Red on First) * P(green on second) P(Red on first, green on second) = 1/7 * 1/4 P(Red on first, green on second) = [B]1/28[/B]

A certain group of woman has a 0.69% rate of red/green color blindness. If a woman is randomly selected, what is the probability that she does not have red/green color blindness? 0.69% = 0.0069. There exists a statistics theorem for an event A that states: P(A) + P(A') = 1 where A' is the event not happening In this case, A is the woman having red/green color blindness. So A' is the woman [U][B][I]not[/I][/B][/U][I] having red/green color blindness[/I] So we have: 0.0069 + P(A') = 1 Subtract 0.0069 from each side, we get: P(A') = 1 - 0.0069 P(A') = [B]0.9931[/B]

A set of 19 scores has a mean of 6.3. A new score of 8 is then included in the data set. What is the new mean? We know the mean formula is: Sum of scores / Number of Scores = Mean We're given mean = 6.3 and number of scores = 19, so we have: Sum of scores / 19 = 6.3 Cross multiply: Sum of scores = 19 * 6.3 Sum of scores = 119.7 Now a new score is added of 8, so we have: Sum of scores = 119.7 + 8 = 127.7 Number of scores = 19 + 1 = 20 So our new mean is: Mean = Sum of scores / Number of Scores Mean = 127.7/20 Mean = [B]6.385[/B] [COLOR=rgb(0, 0, 0)][SIZE=5][FONT=arial][B][/B][/FONT][/SIZE][/COLOR]

Free Addition and Multiplication Tables (Times Tables) Calculator - Shows the color coded addition or multiplication table entries and answer for any 2 numbers 1-15.

All the colors of the rainbow You can find this with the acronym: ROYGBIV [LIST=1] [*][B]Red[/B] [*][B]Orange[/B] [*][B]Yellow[/B] [*][B]Green[/B] [*][B]Blue[/B] [*][B]Indigo[/B] [*][B]Violet[/B] [/LIST] [U]Written as a set, we have 7 elements:[/U] {[B]Red, Orange, Yellow, Green, Blue, Indigo, Violet[/B]}

Belen can make 15 necklaces in 3 1/2 hours. How many can she make in one hour? We set up a proportion of necklaces to time, where n is the number of necklaces Belen can make in 1 hour: 3 & 1/2 = 3.5, so we have: 15/3.5 = n/1 [SIZE=3][FONT=Helvetica][COLOR=rgb(34, 34, 34)] To solve this proportion, we [URL='https://www.mathcelebrity.com/prop.php?num1=15&num2=n&den1=3.5&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']type it in our search engine and we ge[/URL]t: n = [B]4.29 hours[/B][/COLOR][/FONT][/SIZE]

Consider a probability model consisting of randomly drawing two colored balls from a jar containing 2 red and 1 blue balls. What is the Sample Space of this experiment? (assume B= blue and R=red) The sample space is the list of all possible events [LIST] [*]RRB [*]RBR [*]BRR [/LIST]

[COLOR=black]Thanks![/COLOR] [URL='https://www.certificationanswers.com'][COLOR=black]Regards,[/COLOR][/URL]

The formula is: Numerator = One color Denominator - Numerator = Another Color [MEDIA=youtube]mJD2acCpol4[/MEDIA]

Jeremy ran 27 laps on a track that was 1/8 mile long. Jimmy ran 15 laps on a track that as 1/4 mile long. who ran farther [U]Calculate Jeremy's distance:[/U] Distance = Laps * Track length Jeremy distance = 27 * 1/8 Jeremy distance = 27/8 [U]Calculate Jimmy's distance:[/U] Distance = Laps * Track length Jeremy distance = 15* 1/4 Jeremy distance = 15/4 [COLOR=#000000]Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=27/8&frac2=15/4&pl=Compare']fraction comparison calculator[/URL], we see that [B]Jimmy [/B]ran farther[/COLOR]

Free Resistor Color Codes Calculator - Given 3 Band level color codes and a tolerance color chosen, this calculates the resistance in ohms and the tolerance percentage

Free RGB and HEX conversions Calculator - This converts HTML colors from RGB (Red, Green, Blue) to HEX and HEX to RGB (Red, Green, Blue) as well as calculate the Hue.

Sara has a box of candies. In the box there are 8 pink candies, 7 purple candies and 5 blue candies. She takes one candy and records its color. She then puts it back in the box and draws another candy. What is the probability of taking out a pink candy followed by a blue candy? [B][U]Calculate the total number of candies:[/U][/B] Total candies = Pink + Purple + Blue Total candies = 8 + 7 + 5 Total candies = 20 [B][U]Calculate the probability of drawing one pink candy:[/U][/B] P(Pink) = 8/20 Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=8%2F20&frac2=3%2F8&pl=Simplify']fraction reduction calculator[/URL], we get: P(Pink) = 2/5 [B][U]Calculate the probability of drawing one blue candy:[/U][/B] P(Blue) = 5/20 <-- [I]20 options since Sara replaced her first draw[/I] Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F20&frac2=3%2F8&pl=Simplify']fraction reduction calculator[/URL], we get: P(Blue) = 1/4 The problem asks for the probability of a Pink followed by a Blue. Since each event is independent, we multiply: P(Pink, Blue) = P(Pink) * P(Blue) P(Pink, Blue) = 2/5 * 1/4 P(Pink, Blue) = 2/20 Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F20&frac2=3%2F8&pl=Simplify']fraction reduction calculator[/URL], we get: P(Pink, Blue) = [B]1/10 or 10%[/B]

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States that begin with the letter C Our set has 3 elements: [B]{California, Colorado, Connecticut}[/B]

Suppose 5% of a package of balloons are yellow balloons. Then what percentage of balloons are a different color than yellow? Our total possibilities of any color balloon is 100%. So we have: Non-yellow balloons = 100% - yellow balloons percent Non-yellow balloons = 100% - 5% Non-yellow balloons = [B]95%[/B]

Taylor is playing a game using a die and a spinner. The spinner is divided into 4 equal parts with colors green, red, yellow, and purple. Taylor rolls the die and spins the spinner. What is the probability the die shows a 2 and the spinner lands on purple? Probability of rolling a 2 on the die is 1/6 Probability of getting a purple on the spinner is 1/4 Since each event is independent, our joint probability is: P(2 on the die and Purple on the spinner) = P(2 on the die) x P(Purple on the Spinner) P(2 on the die and Purple on the spinner) = 1/6 x 1/4 P(2 on the die and Purple on the spinner) = [B]1/24[/B]

The average height of a family of 6 is 6 feet. After the demise of the mother, the average height remained the same. What is the height of the mother? [LIST] [*]Let the height of the family without the mom be f. Let the height of the mother be m. [*]Averages mean we add the heights and divide by the number of people who were measured. [/LIST] We're given two equations: [LIST=1] [*](f + m)/6 = 6 [*]f/5 = 6 [/LIST] Cross multiplying equation (2), we get: f = 5 * 6 f = 30 Plug f = 30 into equation (1), we get: (30 + m)/6 = 6 Cross multiplying, we get: m + 30 = 6 * 6 m + 30 = 36 To solve this equation for m, we [URL='https://www.mathcelebrity.com/1unk.php?num=m%2B30%3D36&pl=Solve']type it in our search engine[/URL] and we get: m = [B]6[/B] [SIZE=3][FONT=Arial][COLOR=rgb(34, 34, 34)][/COLOR][/FONT][/SIZE]

The Henson family cleaned out all their drawers. They found 47 black pens and 39 blue pens. They also found 6 pens in other colors. How many pens did they find in all? The phrase [I]in all[/I] means we add, so we have: Total pens = Black Pens + Blue Pens + Other color pens Total pens = 47 + 39 + 6 Total pens = [B]92[/B]