Of the 8 students in the chess club, three will represent the club at an upcoming competition. How many different 3-person teams can be formed from the 8 students?

Answers

Answer 1
Answer: I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C(n,r) = n!/(n-r)!r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have

C(8,3) = 8!/(8-3)!3!
C(8,3) = 56

There are 56 3-person teams that can be formed from the 8 students. 

Related Questions

Christiana works at Texas Road House and makes $50 each night in tips plus $7.50 per hour. Write a function rule in function notation where M is the total money she makes and h is the number of hours she works in one night.
A researcher records the time (in seconds) that participants arrive late for a scheduled research study. Assuming these data are normally distributed, which measure of central tendency is most appropriate to describe these data?
The quotient of two numbers is 20. Their sum is 84. What are the two numbers?
How to write 2x+8 in word form
Which statement about square root x-5 - square root x = 5 is true?a) x = –3 is a true solution. b) x = –3 is an extraneous solution. c) x = 9 is a true solution. d) x = 9 is an extraneous solution.

The dataset below represents the population densities per square mile of land area in 15 states in the 2010 U.S. Census. What is the interquartile range?

Answers

Answer:

The Interquartile range is 188

Step-by-step explanation:

Missing Data:

1,19,35,43,49,55,63,94,105,110,175,231,239,351,738

Required

Determine the Interquartile range (IQR)

The given data is ordered already.

First, we need to determine the median

For odd number of data

Median = ½(n + 1)th

In this case, n = 15; so

Median = ½(15 + 1)th

Median = ½(16)th

Median = 8th

This implies that the median is at the 8th position.

So, we have:

1,19,35,43,49,55,63 ----> Lower

(94) ---- Median

105,110,175,231,239,351,738 ---- Upper

Next, we determine the median of the lower and upper sets.

These are called lower quartile (Q1) and upper quartile (Q3) respectively

Lower: 1,19,35,43,49,55,63

Number of data, n = 7

Q1 = ½(n + 1)th

Q1 = ½(7 + 1)th

Q1 = ½(8)th

Q1 = 4th position

From the list of data in the lower set,

Q1 = 43

Upper: 105,110,175,231,239,351,738

Number of data, n = 7

Q3 = ½(n + 1)th

Q3 = ½(7 + 1)th

Q3 = ½(8)th

Q3 = 4th position

From the list of data in the upper set,

Q3 = 231

IQR is then calculated as thus:

IQR = Q3 - Q1

IQR = 231 - 43

IQR = 188

12 more than the quotient of a number and eight

Answers

Answer:

X÷8+12 is how this would look

Answer:

x/8 + 12

Step-by-step explanation:

You are completing ywr math homework of 16 problems. Yo complete 75% before. dinner. How many do yw have left to complete?

Answers

Answer:

4

Step-by-step explanation:

16/4 = 4

25% = 4

50% = 8

75% = 12

100% = 16

Part B: Factor x2 + 10x + 25. Show your work. (3 points)

Answers

x^(2) +10x+25 \n = x^(2) +5x+5x+25 \n =( x^(2) +5x)+(5x+25) \n =x(x+5)+5(x+5) \n =(x+5)(x+5) \n = (x+5)^(2)

8x^3-27=0 how do I solve this



Answers

8x^3 - 27 = 0
x^3 -27 = 0
x^3 = 27
x = 3

Im taking the common core algebra 1 regent and im horrible at math and on top of that i have a really bad math teacher and learned no math this year can some on help he find a site or anything that can help me get at least a 70 (ps. im not a bad student i just hate math my other regents i got high 90's)

Answers

This site can help you.