TutorMe homepage

SIGN IN

Start Free Trial

Mark v.

B.S. in Mathematics and Computer Science

Tutor Satisfaction Guarantee

Discrete Math

TutorMe

Question:

Let there be a positive integer n. Show that for any set of n consecutive integers there is one divisible by n.

Mark v.

Answer:

Let c, c+1, c+2, c+3, ... c+n-1 be the integers in this sequence. If r = (c) mod n, then either r = 0 or c < c+n-r ≤ c+n-1, and therefore in the sequence. Since (n-r) mod n = n-r, (c+n-r) mod n =0. Therefore, there must exist one integer in the sequence that is divisible by n.

Statistics

TutorMe

Question:

Bob and Alice are playing a game of dice throwing. Bob found a eight-sided dice, where in Alice only found two four-sided dice. What are the odds that the sum of Alice's results are greater than Bob's?

Mark v.

Answer:

Each of Bob's twelve possible out comes; {1, 2, 3, 4, 5, 6, 7, 8} are equally likely to occur, at a 1/8th probability. Alice's outcomes; {2, 3, 4, 5, 6, 7, 8} have a {1/16th, 2/16th, 3/16th, 4/16th, 3/16th, 2/16th, 1/16th} probability, respectively. To determine the odds of Alice rolling greater the Bob, we multiply each of Bob's results by the chance Alice rolls greater than that result. There is a 1/8th chance of Bob rolling a 1, and a 16/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 2, and a 15/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 3, and a 13/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 4, and a 10/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 5, and a 6/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 6, and a 3/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 7, and a 1/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 8, and a 0/16th chance Alice rolls greater. Multiplying those chances together, the total probability p that Alice rolls greater then Bob is as follows; p=(1/8)*(16/16)+(1/8)*(15/16)+(1/8)*(13/16)+(1/8)*(10/16)+(1/8)*(6/16)+(1/8)*(3/16)+(1/8)*(1/16)+(1/8)*(0/16) p=(1/8)*(1/16)*(16+15+13+10+6+3+1+0) p=64/128=0.5 There is a 50% chance of Alice's two four-sided dice roll a higher total than Bob's one eight-sided dice

Algebra

TutorMe

Question:

Given that: x^2 +y^2 = 25 and 2x = y. What are are all possible values for x and y.

Mark v.

Answer:

Since we know that y = 2x, from to the second equation, we can replace the y in the first equation with 2x. Then x^2+(2x)^2=25 can be simplified to x^2+4(x^2)=25 thanks to the commutative property of multiplication. It can be further simplified to 5(x^2)=25 by the distributive property. Dividing both sides by 5 will leave us with x^2=5, and if we take the square root of both sides, we have x=±√5. So, given that 2x=y, the possible values for x and y are √5 and 2√5, and -√5 and -2√5.

Send a message explaining your

needs and Mark will reply soon.

needs and Mark will reply soon.

Contact Mark

Ready now? Request a lesson.

Start Session

FAQs

What is a lesson?

A lesson is virtual lesson space on our platform where you and a tutor can communicate.
You'll have the option to communicate using video/audio as well as text chat.
You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.

How do I begin a lesson?

If the tutor is currently online, you can click the "Start Session" button above.
If they are offline, you can always send them a message to schedule a lesson.

Who are TutorMe tutors?

Many of our tutors are current college students or recent graduates of top-tier universities
like MIT, Harvard and USC.
TutorMe has thousands of top-quality tutors available to work with you.

Made in California

© 2018 TutorMe.com, Inc.