2 buses leave at 5:30am, 1 comes every 18 minutes and one comes every 15 minutes when will they both come at the same time again We want the Least Common Multiple (LCM) of 15 and 18. LCM(15, 18) Enter this into the [URL='http://www.mathcelebrity.com/gcflcm.php?num1=15&num2=18&num3=&pl=LCM']search engine[/URL], and we get: [B]90 minutes[/B]

2 traffic lights are turned on at the same time. 1 blinks every 4 seconds and. the other blinks every 6 seconds. In 60 seconds how many times will they blink at the same time? We want the [URL='https://www.mathcelebrity.com/gcflcm.php?num1=4&num2=6&num3=&pl=LCM']least common multiple of 4 and 6[/URL] which is 12. So ever 12 seconds, both lights blink together: [LIST=1] [*]12 [*]24 [*]36 [*]48 [*]60 [/LIST] So our answer is [B]5 times[/B]

A 100 point test contains a total of 20 questions. The multiple choice questions are worth 3 points each and short response questions are worth 8 points each. Write a system of linear equations that represents this situation Assumptions: [LIST] [*]Let m be the number of multiple choice questions [*]Let s be the number of short response questions [/LIST] Since total points = points per problem * number of problems, we're given 2 equations: [LIST=1] [*][B]m + s = 20[/B] [*][B]3m + 8s = 100[/B] [/LIST] We can solve this system of equations 3 ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+s+%3D+20&term2=3m+%2B+8s+%3D+100&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+s+%3D+20&term2=3m+%2B+8s+%3D+100&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+s+%3D+20&term2=3m+%2B+8s+%3D+100&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get: [B]m = 12, s = 8[/B]

a bell ring every 15 seconds another bell ring 30 seconds.at 3:00 pm the 2 bells ring simultaneously.at what time will the bells ring again at the same time The [URL='https://www.mathcelebrity.com/gcflcm.php?num1=15&num2=30&num3=&pl=GCF+and+LCM']Least Common Multiple (LCM)[/URL] of 15 and 30 is 30: Therefore, 30 seconds from now, 3:00, is when the 2 bells will ring simultaneously. We add 30 seconds to 3:00 and get: 3:00 and 30 seconds.

A bell rings every 18 seconds, while another bell rings every 60 seconds. At 5:00 pm the two ring simultaneously. At what time will be the bell ring again at the same time. We want the Least Common Multiple of 18 and 60. Using our [URL='https://www.mathcelebrity.com/gcflcm.php?num1=18&num2=60&num3=&pl=GCF+and+LCM']least common multiple of 18 and 60[/URL] is [B]180 [/B] 180/18 = 10 (18 second periods) 180/60 = 3 (60 second periods) 180 seconds = 3 minutes So the next time the bells ring simultaneously is 5:00 + 3 = [B]5:03 pm[/B]

A comet passes earth every 70 years. another comet passes earth every 75 years of both comets pass earth this year how many years will it be before they pass on the same year again. We want the least common multiple of (70, 75). [URL='https://www.mathcelebrity.com/gcflcm.php?num1=70&num2=75&num3=&pl=GCF+and+LCM']Using our LCM calculator[/URL], we find the answer is [B]1,050 years[/B]

A company has 12,600 employees. Of these, 1/4 drive alone to work, 1/6 car pool, 1/8 use public transportation, 1/9 cycle, and the remainder use other methods of transportation. How many employees use each method of transportation? Find the remainder fraction: Remainder = 1 - (1/4 + 1/6 + 1/8 + 1/9) The least common multiple of 4, 6, 8, 9 is 72. So we divide 72 by each fraction denominator to get our multiplier: 1/4 = 18/72 1/6 = 12/72 1/8 = 9/72 1/9 = 8/72 Add those all up: (18 + 12 + 9 + 8)/72 47/72 Now subtract the other methods out from 1 to get the remainder of who use other methods: Remainder = 1 - 47/72 Since 1 = 72/72, we have: (72 - 47)/72 [B]25/72[/B]

A computer randomly generates a whole number from 1 to 25. Find the probability that the computer generates a multiple of 5 [URL='https://www.mathcelebrity.com/factoriz.php?num=25&pl=Show+Factorization']Multiples of 5[/URL]: {1, 5, 25} So we have the probability of a random number multiple of 5 is [B]3/25[/B]

A farmer is taking her eggs to the market in a cart, but she hits a pothole, which knocks over all the containers of eggs. Though she is unhurt, every egg is broken. So she goes to her insurance agent, who asks her how many eggs she had. She says she doesn't know, but she remembers somethings from various ways she tried packing the eggs. When she put the eggs in groups of two, three, four, five, and six there was one egg left over, but when she put them in groups of seven they ended up in complete groups with no eggs left over. What can the farmer figure from this information about the number of eggs she had? Is there more than one answer? We need a number (n) that leaves a remainder of 1 when divided by 2, 3, 4, 5, 6 but no remainder when divided by 7. 217 + 84 = [B]301[/B]. Other solutions are multiples of 3 x 4 x 5 x 7, but we want the lowest one here.

A first number plus twice a second number is 10. Twice the first number plus the second totals 35. Find the numbers. [U]The phrase [I]a number[/I] means an arbitrary variable[/U] A first number is written as x A second number is written as y [U]Twice a second number means we multiply y by 2:[/U] 2y [U]A first number plus twice a second number:[/U] x + 2y [U]A first number plus twice a second number is 10 means we set x + 2y equal to 10:[/U] x + 2y = 10 [U]Twice the first number means we multiply x by 2:[/U] 2x [U]Twice the first number plus the second:[/U] 2x + y [U]Twice the first number plus the second totals 35 means we set 2x + y equal to 35:[/U] 2x + y = 35 Therefore, we have a system of two equations: [LIST=1] [*]x + 2y = 10 [*]2x + y = 35 [/LIST] Since we have an easy multiple of 2 for the x variable, we can solve this by multiply the first equation by -2: [LIST=1] [*]-2x - 4y = -20 [*]2x + y = 35 [/LIST] Because the x variables are opposites, we can add both equations together: (-2 + 2)x + (-4 + 1)y = -20 + 35 The x terms cancel, so we have: -3y = 15 To solve this equation for y, we [URL='https://www.mathcelebrity.com/1unk.php?num=-3y%3D15&pl=Solve']type it in our search engine[/URL] and we get: y = [B]-5 [/B] Now we substitute this y = -5 into equation 2: 2x - 5 = 35 To solve this equation for x, we[URL='https://www.mathcelebrity.com/1unk.php?num=2x-5%3D35&pl=Solve'] type it in our search engine[/URL] and we get: x = [B]20[/B]

A light flashes every 2 minutes a, second light flashes every 7 minutes, and a third light flashes every 8 minutes. If all lights flash together at 8 P.M., what is the next time of day they will all flash together [URL='https://www.mathcelebrity.com/gcflcm.php?num1=2&num2=7&num3=8&pl=LCM']We use our least common multiple calculator[/URL] to see when the 3 numbers have a common multiple: LCM of (2, , 8) = 56 minutes So this means we add 56 minutes to 8:00 P.M. and we get [B]8:56 P.M.[/B]

a lighthouse blinks every 12 minutes. A second lighthouse blinks every 10 minutes if they both blink at 8:10 P.M at what time will they next blink together We want the least common multiple of (10, 12). This will be the next time each number times a multiple equals the same number. [URL='https://www.mathcelebrity.com/gcflcm.php?num1=10&num2=12&num3=&pl=GCF+and+LCM']Typing in LCM 10,12 into our search engine[/URL], we get: 60 So if we start at 8:10, and 60 minutes later is when both lighthouses blink. 60 minutes equals 1 hour. So we add 1 hour to 8:10, we have [B]9:10[/B]

A lighthouse blinks every 12 minutes.A second lighthouse blinks every 10 minutes.If they both blink at 8:10 P.M., at what time will they next blink together? We want to know the least common multiple, so that 12 and 10 intervals meet again.[URL='https://www.mathcelebrity.com/gcflcm.php?num1=10&num2=12&num3=&pl=GCF+and+LCM'] We type in LCM(10,12) into our search engine[/URL] and we get 60. 60 minutes is 1 hour, so we add this to 8:10 to get [B]9:10[/B]

A math test is worth 100 points and has 38 problems. Each problem is worth either 5 points or 2 points. How many problems of each point value are on the test? Let's call the 5 point questions m for multiple choice. Let's call the 2 point questions t for true-false. We have two equations: [LIST=1] [*]m + t = 38 [*]5m + 2t = 100 [/LIST] Rearrange (1) to solve for m - subtract t from each side: 3. m = 38 - t Now, substitute (3) into (2) 5(38 - t) + 2t = 100 190 - 5t + 2t = 100 Combine like terms: 190 - 3t = 100 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=190-3t%3D100&pl=Solve']equation solver[/URL], we get [B]t = 30[/B]. Plugging t = 30 into (1), we get: 30 + t = 38 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=m%2B30%3D38&pl=Solve']equation solver[/URL] again, we get [B]m = 8[/B]. Check our work for (1) 8 + 30 = 38 <-- Check Check our work for (2) 5(8) + 2(30) ? 100 40 + 60 ? 100 100 = 100 <-- Check You could also use our [URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+t+%3D+38&term2=5m+%2B+2t+%3D+100&pl=Cramers+Method']simultaneous equations calculator[/URL]

A pair of numbers has an HCF (Highest Common Factor) of 3, and an LCM (Lowest Common Multiple) of 45 . If one of the numbers in the pair is 15 , what is the other number? [LIST=1] [*]Prime Factorization for 15 is 3 * 5 [*]Prime Factorization for 9 is 3 * 3 [*]LCM of (9, 15) = 35 [/LIST] [URL='https://www.mathcelebrity.com/gcflcm.php?num1=9&num2=15&num3=&pl=GCF+and+LCM']Check out this link here to see the details[/URL]

A test has twenty questions worth 100 points . The test consist of true/false questions worth 3 points each and multiple choice questions worth 11 points each . How many multiple choice questions are on the test? Set up equations where t = true false and m = multiple choice: [LIST=1] [*]t + m = 20 [*]3t + 11m = 100 [/LIST] Use our [URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=t+%2B+m+%3D+20&term2=3t+%2B+11m+%3D+100&pl=Cramers+Method']simultaneous equation calculator[/URL]: [B]t = 15, m = 5[/B]

A test has twenty questions worth 100 points total. the test consists of true/false questions worth 3 points each and multiple choice questions worth 11 points each. How many true/false questions are on the test? Let m be the number of multiple choice questions and t be the number of true/false questions. We're given: [LIST=1] [*]m + t = 20 [*]11m + 3t = 100 [/LIST] We can solve this system of equations 3 ways below: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the following answers: [LIST] [*][B]m = 5[/B] [*][B]t = 15[/B] [/LIST] Check our work in equation 1: 5 + 15 ? 20 [I]20 = 20[/I] Check our work in equation 2: 11(5) + 3(15) ? 100 55 + 45 ? 100 [I]100 = 100[/I]

A test has twenty questions worth 100 points. The test consists of True/False questions worth 3 points each and multiple choice questions worth 11 points each. How many multiple choice questions are on the test? Let the number of true/false questions be t. Let the number of multiple choice questions be m. We're given two equations: [LIST=1] [*]m + t = 20 [*]11m + 3t = 100 [/LIST] We have a set of simultaneous equations. We can solve this using 3 methods: [LIST=1] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=1m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=1m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=1m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we pick, we get the same answer: [LIST] [*][B]m = 5[/B] [*][B]t = 15[/B] [/LIST]

A toad croaks every 8seconds and a frog croaks every 6 seconds .They both croak at the same .After how many seconds will they next croak at the same time again. We want the least common multiple of 8 and 6. We type in [URL='https://www.mathcelebrity.com/gcflcm.php?num1=6&num2=8&num3=&pl=GCF+and+LCM']LCM(6, 8) into our search engine[/URL] and we get [B]24[/B]

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Amar goes to the dance class every fourth day. Karan goes to the dance class every fifth day. Both met at the dance class today. After how many days will they meet at the dance class again? We want the least common multiple of 4 and 5. We type in [URL='https://www.mathcelebrity.com/gcflcm.php?num1=4&num2=5&num3=&pl=GCF+and+LCM']LCM(4,5)[/URL] into our search engine and we get [B]20. So 20 days from now, Amar and Karen will meet again.[/B]

An air horn goes off every 48 seconds,another every 80 seconds. At 5:00 pm the two go off simultaneously. At what time will the air horns blow again at the same time? We want to find the least common multiple of (48, 80). So we [URL='https://www.mathcelebrity.com/gcflcm.php?num1=48&num2=80&num3=&pl=GCF+and+LCM']type this in our search engine[/URL], and we get: 240. So 240 seconds is our next common meeting point for each air horn. When we [URL='https://www.mathcelebrity.com/timecon.php?quant=240&pl=Calculate&type=second']type 240 seconds into our search engine[/URL], we get 4 minutes. We add the 4 minutes to the 5:00 pm time to get [B]5:04 pm[/B].

An indoor water park has two giant buckets that slowly fill with 1000 gallons of water before dumping it on the people below. One bucket dumps water every 18 minutes. The other bucket dumps water every 21 minutes. It is currently 1:15 P.M. and both buckets dumped water 5 minutes ago. Find the next two times that both buckets dump water at the same time. We want to find the Least Common Multiple between 18 minutes and 21 minutes. This shows us when both bucket dumping cycles happen simultaneously. So we[URL='https://www.mathcelebrity.com/gcflcm.php?num1=18&num2=21&num3=&pl=GCF+and+LCM'] type in LCM(18,21) into our search engine and we get[/URL]: LCM(18, 21) = 126 This means, in 126 minutes, both buckets will dump water. Since 60 minutes is in an hour, we find out how many full hours we have. To find the full hours and remainder, we [URL='https://www.mathcelebrity.com/modulus.php?num=126mod60&pl=Calculate+Modulus']type in 126 mod 60[/URL] into our search engine and we get: 6. This means 126 minutes is 2 hours and 6 minutes. Find the next bucket dumping time: [LIST=1] [*]We start at 1:15 PM [*]Add 2 hours and we get 3:15 PM [*]Add 6 minutes and we get [B]3:21 PM[/B] [/LIST]

Anna painted 1/6 of a wall, Eric painted 1/5 of the wall, and Meadow painted 1/4 of the wall. There are now 3910 square feet left to paint. How many square feet did Anna paint? [URL='https://www.mathcelebrity.com/gcflcm.php?num1=4&num2=5&num3=6&pl=LCM']Using 60 as a common denominator through least common multiple[/URL], we get: 1/6 = 10/60 1/5 = 12/60 1/4 = 15/60 10/60 + 12/60 + 15/60 = 37/60 Remaining part of the wall is 60/60 - 37[B]/[/B]60 = 23/60 3910/23 = 170 for each 1/60 of a wall Anna painted 1/6 or 10/60 of the wall. So we multiply 170 * 10 = [B]1,700 square feet[/B]

Ben visits the park every 2 days and goes to the library every 5 days. If Ben gets to do both of these today, how many days will pass before Ben gets to do them both on the same day again? To find this, we want the least common multiple (LCM) of 2 and 5. We [URL='https://www.mathcelebrity.com/gcflcm.php?num1=2&num2=5&num3=&pl=GCF+and+LCM']type LCM(2,5) into our search engine[/URL] and we get: [B]10 days [/B] We check our work: 2 days * 5 visits = 10 days 5 days * 2 visits = 10 days

If a seed is planted, it has a 90% chance of growing into a healthy plant. If 12 seeds are planted, what is the probability that exactly 4 don't grow? Im seriously confused is it like u multiple the amount of the (0.90) and multiple (0.30) by power depends how any they r right?

Blake and Tatsu are each assigned a paper for a class they share. Blake decides to write 4 pages at a time while Tatsu decides to write 7 pages at a time. If they end up writing the same number of pages, what is the smallest number of pages that the papers could have had? We want the least common multiple of 4 and 7, written as LCM(4, 7). Using our [URL='http://www.mathcelebrity.com/gcflcm.php?num1=4&num2=7&num3=&pl=LCM']LCM Calculator[/URL], we get: LCM(4, 7) = [B]28 pages[/B]

Caleb has a complicated and difficult research paper due soon. What should he do to keep from feeling overwhelmed and procrastinating? A. work on the paper every day but save the bulk of the work for the night before it's due B. break down the paper into several small steps and start with the smallest one C. write down the deadline for the paper where he can see it every day so he doesn't forget D. work on the hardest parts of the paper first and take multiple breaks until he's finished Caleb wants to avoid both overwhelm and procrastination. Let's review each option: [LIST] [*]A is out because saving a majority of the work will cause overwhelm [U]and[/U] demonstrates procrastination [*]B is a good option as small steps reduce overwhelm [*]C looks nice on paper, but will he follow through with seeing the deadline everyday? [*]D is a good option as well. Finishing the tough parts first makes the rest of the journey seem like a downhill cruise [/LIST] Based on these, I'd take [B]B or D[/B]

Every 10 customers receive a soda. Every 9 customers receive a hot dog. There are 450 customers. How many received both a soda and hot dog? Fine the Least Common Multiple of 9 and 10 using our [URL='http://www.mathcelebrity.com/gcflcm.php?num1=9&num2=10&num3=&pl=LCM']LCM Calculator[/URL]: LCM(9, 10) = 90. So every 90 customers receives a soda [U][B]and[/B][/U] a hot dog. We have 90, 180, 270, 360, 450, so [B]5 customers[/B] receive both.

This is a least common multiple problem. [URL='http://www.mathcelebrity.com/gcflcm.php?num1=6&num2=8&num3=&pl=LCM']The least common multiple of 6 and 8 is 24[/URL] So every 24th person, less than or equal to 329 receives both a soda [U]and[/U] a hot dog. Using our multiples calculator, we find there are [URL='http://www.mathcelebrity.com/multiple.php?num=24&pl=Multiplication+Multiples']13 multiples of 24 less than or equal to 329[/URL]. 24,48,72,96,120,144,168,192,216,240,264,288,312

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Express cos4θ and sin4θ in terms of sines and cosines of multiples of θ. Using a trignometric identity: cos (2θ) = cos^2(θ) - sin^2(θ) Since 4θ = 2*2θ, so we have: [B]cos(4θ) = cos^2(2θ) - sin^2(2θ)[/B] Using another trignometric identity, we have: sin(2θ) = 2 sin(θ) cos(θ) Since 4θ = 2*2θ, so we have: [B]sin(4θ) = 2 sin(2θ) cos(2θ)[/B]

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Jinas final exam has true/false questions, worth 3 points each, and multiple choice questions, worth 4 points each. Let x be the number of true/false questions she gets correct, and let y be the number of multiple choice questions she gets correct. She needs at least 76 points on the exam to get an A in the class. Using the values and variables given, write an inequality describing this. At least means greater than or equal to, so we have: [B]3x + 4y >= 76[/B]

Max sneezes every 5 minutes, Lina coughs every 6 minutes, and their dog barks every 3 minutes. If there was sneezing, barking, and coughing at 3:15 PM, when is the next time that these three sounds will happen simultaneously? To find the next time the sounds happen simultaneously, we want to find the Least Common Multiple (LCM). [URL='https://www.mathcelebrity.com/gcflcm.php?num1=3&num2=5&num3=6&pl=LCM']Using our LCM Calculator[/URL], we find the least common multiple of 3, 5, and 6 is 30. The least common multiple gives us a common time where each sound reaches a "cycle". [LIST] [*]Dog: A bark every e minutes means the dog has 10 barks, with the 10th bark at 30 minutes after 3:15 [*]Max: A sneeze every 5 minutes means he has 6 sneezes, with the 6th sneeze at 30 minutes after 3:15 [*]Lisa: A cough every 6 minutes means she has 5 coughs, with the 5th cough at 30 minutes after 3:15 [/LIST] 30 minutes after 3:15 means we have: 3:15 + 30 = [B]3:45 PM[/B]

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Pete ate 1/2 a pizza, Carol ate 1/3, and Joe ate 1/4. How much pizza was eaten? Total pizza eaten is: 1/2 + 1/3 + 1/4 We need a common denominator. List out factors: 2: 1, 2, 4, 6, 8, 10, 12 3: 1, 3, 6, 9, 12 4: 1, 4, 8, 12 12 is our least common multiple: 1/2 = 6/12 1/3 = 4/12 1/4 = 3/12 So our addition becomes: (6 + 4 + 3)/12 = 13/12 of a pizza

Some hot dogs come in packages of 8. Why would a baker of hot dog buns package 7 hot dog buns to a package For customers that like to have matching hot dogs and buns, consider this scenario. For the first round, you have one extra hot dog. Now you buy a hot dog buns package. You're over 6 buns. This continues... We want to see when packaging and hot dogs math. Find the least common multiple of 7 and 8 so that packages match. [URL='https://www.mathcelebrity.com/gcflcm.php?num1=7&num2=8&num3=&pl=LCM']LCM(7, 8[/URL][I][URL='https://www.mathcelebrity.com/gcflcm.php?num1=7&num2=8&num3=&pl=LCM']) [/URL]= 56[/I]

Suppose that the manager of the Commerce Bank at Glassboro determines that 40% of all depositors have a multiple accounts at the bank. If you, as a branch manager, select a random sample of 200 depositors, what is the probability that the sample proportion of depositors with multiple accounts is between 35% and 50%? [URL='http://www.mathcelebrity.com/proportion_hypothesis.php?x=50&n=+100&ptype==&p=+0.4&alpha=+0.05&pl=Proportion+Hypothesis+Testing']50% proportion probability[/URL]: z = 2.04124145232 [URL='http://www.mathcelebrity.com/proportion_hypothesis.php?x=+35&n=+100&ptype==&p=+0.4&alpha=+0.05&pl=Proportion+Hypothesis+Testing']35% proportion probability[/URL]: z = -1.02062072616 Now use the [URL='http://www.mathcelebrity.com/zscore.php?z=p%28-1.02062072616

There are two bells in the school. Bell A rings every 2 minutes. Bell B rings every 3 minutes. If both bells ring together at 8.02 p.m., when will they ring together again? Using our[URL='http://www.mathcelebrity.com/gcflcm.php?num1=2&num2=3&num3=&pl=LCM'] least common multiple calculator,[/URL] we find the LCM(2, 3) = 6. Which means the next time both bells ring together is 6 minutes from now. 8:02 p.m. + 6 minutes = [B]8:08 p.m. [MEDIA=youtube]4dfLeHimP8Q[/MEDIA][/B]

True or False (a) The normal distribution curve is always symmetric to its mean. (b) If the variance from a data set is zero, then all the observations in this data set are identical. (c) P(A AND A^c)=1, where A^c is the complement of A. (d) In a hypothesis testing, if the p-value is less than the significance level α, we do not have sufficient evidence to reject the null hypothesis. (e) The volume of milk in a jug of milk is 128 oz. The value 128 is from a discrete data set. [B](a) True, it's a bell curve symmetric about the mean (b) True, variance measures how far a set of numbers is spread out. A variance of zero indicates that all the values are identical (c) True. P(A) is the probability of an event and P(Ac) is the complement of the event, or any event that is not A. So either A happens or it does not. It covers all possible events in a sample space. (d) False, we have sufficient evidence to reject H0. (e) False. Volume can be a decimal or fractional. There are multiple values between 127 and 128. So it's continuous.[/B]

What is the Least Common Multipole of 3,4,5? Using our l[URL='https://www.mathcelebrity.com/gcflcm.php?num1=3&num2=4&num3=5&pl=LCM']east common multiple calculator[/URL], we get: [B]60[/B]

What is the sum of four consecutive multiples of 5? First number = n Second number = n + 5 Third number = n + 10 Fourth number = n + 15 Add them together: (n + n + n + n) + (5 + 10 + 15) [B]4n + 30[/B]

x is a multiple of 6 and 1 ≤ x ≤ 16. We want multiples of 6 between 1 and 16. We start with 6. Another multiple of 6 is 12 The next multiple of 6 is 18, which is out side the range of 1 ≤ x ≤ 16. So our number set is [B]x = {6, 12}[/B]