Mathematics, 01.08.2019 12:30 gaby8282

Astarting lineup in basketball consists of two guards, two forwards, and a center. (a) a certain college team has on its roster four centers, five guards, three forwards, and one individual (x) who can play either guard or forward. how many different starting lineups can be created? [hint: consider lineups without x, then lineups with x as guard, then lineups with x as forward.] 240 lineups

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Answer from: allieballey0727

1 . Consider lineups without x. You have to select

2 guards from 5 quards;2 forwards from 3 forwards;1 center from 4 centers.

It can be made in

$C_5^2\cdot C_3^2\cdot C_4^1=10\cdot 3\cdot 4=120$ different ways.

2. Consider lineups with x as guard. You have to select

2 guards from 6 quards (one of them must be x);2 forwards from 3 forwards;1 center from 4 centers.

It can be made in

$C_5^1\cdot C_3^2\cdot C_4^1=5\cdot 3\cdot 4=60$ different ways.

3. Consider lineups with x as forward. You have to select

2 guards from 5 quards;2 forwards from 4 forwards (one of them must be x);1 center from 4 centers.

It can be made in

$C_5^2\cdot C_3^1\cdot C_4^1=10\cdot 3\cdot 4=120$ different ways.

Therefore, the total number of different lineups is

120+60+120=300.

Answer from: stuckonquestions

300

Justification:

Follow the hint:

1) Line ups without x:

Number of guards to pick: 2 of 5 => C5,2
Number of forwards to pick: 2 of 3 => C3,2
Number of centers to pick 1 of 4 => C4,1

Number of possible combinations: C5,2 * C3,2 * C4,1

The formula for Cm,n is: Cm,n = m! / [n! * (m-n)! ]

=>

C5,2 = 5! / [2! (5-3)! ] = 5*4 / 2 = 10

C3,2 = 3! / [2! (3-1! ] = 3

C4,1 = 4! / [1!(4-1)!] = 4

=> 10*3*4 = 120 different starting lineups without x

2) Lineups with x as guard

You only need to pick one more guard

number of guards to pick: 1 of 5 => C5,1
number of forwards to pick: 2 of 3 => C3,2
number of centers to pick: 4 => 1 of 4 => C4,1

Number of possible combinations: C5,1 * C3,2 * C4,1 = 5 * 3 * 4 = 60

3) Lineups with x as a forward,

You only need to pick one more forward:

Number of guards to pick: 2 of 5 => C5,2
Number of forwards to pick: 1 of 3 => C3,1
Number of centers to pick: 1 of 4 => C4,1

.Number of possible combinations: C5,2 * C3,1 * C4,1 = 10 * 3 * 4 = 120

4) Total number of different lineups:

Number of lineups without x + number of lineups with x as guard + number of lineups with x as forward = 120 + 60 + 120 = 300

Answer from: Quest

the answer

b. they are parallel and will never intersect.

Answer from: Quest

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation:

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