2 adults, 6 students
4 adults, 3 students
6 adults
9 students
4 combinations
4
7/128, or about 5.5% The student has a 1/2 probability of getting each question correct. The probability that he passes is the probability that he gets 10 correct+probability that he gets 9 correct+probability that he gets 8 correct: P(passes)=P(10 right)+P(9 right)+P(8 right)=[(1/2)^10]+[(1/2)^10]*10+[(1/2)^10]*Combinations(10,2)=[(1/2)^10](1+10+45)=56/1024=7/128.
The student is calculating the average or mean speed of the ball.
because they are tired
P = (6!)/(6-4)!4!=15
Since there are only two options for the answer, on average the student will answer half of the answers correctly.
This is a combinations question. The answer is C(10,8), the number of combinations of 10 things taken 8 at a time, which also equals the number of combinations of 10 things taken 2 at a time. You can imagine why these are the same: choosing 8 questions to answer out of ten is the same thing as not choosing 2 questions to answer out of ten. C(10,2) = (10*9)/(2*1) = 45. There are 45 different choices the student can make of which eight questions to answer.
Any combination of 5 students leaves one student out. Since there are 5 possible students to leave out, the number of combinations of all but one student is 5.
Ludwig van Beethoven composed "Fur Elise" for Therese Malfatti, a student for whom he had a shine and who he hoped to marry.
There are 84 different combinations possible for the committee of 6, taken from4 students and 5 teachers.1.- The committee with 4 students has 4C4 number of combinations of 4 students out of 4 and 5C2 number of combinations of 2 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 4 students and 2 teachers.2.- The committee with 3 students has 4C3 number of combinations of 3 students out of 4 and 5C3 number of combinations of 3 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 3 students and 6 teachers.3.- The committee with 2 students has 4C2 number of combinations of 2 studentsout of 4 and 5C4 number of combinations of 4 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 2 students and 4 teachers.4.- The committee with 1 student has 4C1 number of combinations of 1 student out of 4 students and 5C5 number of combinations of 5 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 1 student and 5 teachers.We now add up all possible combinations:4C4∙5C2 + 4C3∙5C3 + 4C2∙5C4 + 4C1∙5C5 = 1(10) + 4(10) +6(5) + 4(1) = 84There are 84 different combinations possible for the committee of 6, taken from4 students and 5 teachers.[ nCr = n!/(r!(n-r)!) ][ n! = n(n-1)(n-2)∙∙∙(3)(2)(1) ]
Therese Bissen Bard has written: 'Student assistants in the school library media center' -- subject(s): Employees, Instructional materials centers, School libraries, Student library assistants, Volunteer workers in libraries 'John and Mollie Bissen in their time' -- subject(s): Farm life, Biography 'John and Mollie Bissen in their time'
Today in homeroom, the anxious students will receive their locker combinations and a student handbook.The school colors are combinations of green, white, and black.
How are traditional algorithms different from student-invented strategy
A student is a pupil. They are not different because if you call someone pupil they will think that you are calling them a student. So if you are going to call someone pupil all ways rember to think of that they are student
A student version sells for around 400 dollars while a professional version sells for around 800 dollars but it has many more features than the student version.
The average college student spends five dollars a day on coffee. During a typical school week this amounts to twenty-five dollars.
$55,460 dollars for fifty months
Tell your dad to donate millions of dollars to Harvard.