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satela [25.4K]
1 year ago
8

A computer programming team has 17 members. (a) How many ways can a group of nine be chosen to work on a project? As in Example

9.5.4, since the set of people in a group is a subset of the set of people on the team, the answer is (b) Suppose nine team members are women and eight are men. (1) How many groups of nine can be chosen that contain five women and four men? As in Example 9.5.7a, think of forming the group as a two-step process, where step 1 is to choose the women and step 2 is to choose the men. The answer is (ii) How many groups of nine can be chosen that contain at least one man? As in Example 9.5.7b, consider the relationship between the set of groups that consist entirely of women and the set of groups with at least one man. This thinking leads to the conclusion that the number of groups with at least on man is (iii) How many groups of nine can be chosen that contain at most three women? A group of nine that contains at most three women can contain no women, one woman, two women, or three women. So, the number of groups of nine that contain at most three women is (c) Suppose two team members refuse to work together on projects. How many groups of nine can be chosen to work on a project? In a similar way as in Example 9.5.6, let A and B be the two team members who refuse to work together in a group. Thinking about the number of groups that contain A and not B, B and not A, and neither A nor B leads to the conclusion that the total number of groups of that can be chosen to work on the project is (d) Suppose two team members insist on either working together or not at all on projects. How many groups of nine can be chosen to work on a project? As in Example 9.5.5, let A and B be the two team members who insist on working in a group together or not at all. How many groups contain both A and B? How many groups contain neither A nor B?
Mathematics
1 answer:
Natali5045456 [20]1 year ago
7 0

The number of groups that can be chosen from the team of 17 depends on the given conditions of the selection.

The correct responses are;

(a) The number of ways of choosing groups of 9 from the 17 members are 24,310 ways.

(b) (i) The number of groups of 9 containing 5 women and 4 men are 8,820 groups.

(ii) The number of groups of 9 that contain at least one man is 24,309 groups.

(iii) The number of groups of 9 that contain at most 3 women are 2,649 groups.

(c) If two team members refuse to work together on projects, the number of groups of 9 that can be chosen are 17,875 groups.

(d) If two team members must work together or not at all the number of groups of 9 that can be chosen are 1,430 groups.

Given,

The combination of n objects taking r at a time is given as follows

ₙCr = (ⁿr)

(a) The number of ways of choosing groups of 9 from the 17 members is given as follows;

Number of ways of choosing groups of 9 = (17, 9)  = 24,310 ways.

(b) (i) Number of women = 9

Number of men = 8

Number of groups of 9 containing 5 women and 4 men = (9, 5) x (8, 4)

(9, 5) x (8, 4) = 126 x 70 = 8820

Number of groups of 9 containing 5 women and 4 men = 8,820 groups.

(ii) The number of groups that contain at least one man is (17, 9) - (9, 9)

(17, 9) - (9, 9) = 24,310 - 1 = 24,309

The number of groups that contain at least one man are 24,309 groups.

(iii) The number of groups of 9 that contain at most 3 women if given as follows;

(8, 8) × (9, 1) + (8, 7) × (9, 2) + (8, 6) × (9, 3) = 2649

The number of groups of 9 that contain at most 3 women are 2,649 groups.

(c) Given that two team members refuse to work together on projects, the number of groups of 9 that can be chosen are (15, 9) + 2 × (15, 8)

(15, 9) + 2 × (15, 8) = 5,005 + 2 × 6,435 = 17,875

(d) Two team members must work together or not at all is (15, 7) - (15, 9)

(15, 7) - (15, 9) = 6,435 - 5,005 = 1,430

The number of groups of 9 that can be chosen two team members must work together or not at all are 1,430 groups.

Learn more about number of ways here;

brainly.com/question/25310371

#SPJ4

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\bold{\huge{\underline{\pink{ Solution }}}}

<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>

  • <u>Rectangle </u><u>1</u><u> </u><u> </u><u>has </u><u>length </u><u>x</u><u> </u><u>and </u><u>width </u><u>y</u>
  • <u>Rectangle</u><u> </u><u>2</u><u> </u><u>is </u><u>made </u><u>by </u><u>multiplying </u><u>each </u><u>dimensions </u><u>of </u><u>rectangle </u><u>1</u><u> </u><u>by </u><u>a </u><u>factor </u><u>of </u><u>k </u>
  • <u>Where</u><u>, </u><u>k </u><u>></u><u> </u><u>0</u><u> </u><u> </u>

<h3><u>Answer </u><u>1</u><u> </u><u>:</u><u>-</u></h3>

Yes, The rectangle 1 and rectangle 2 are similar .

<h3><u>According </u><u>to </u><u>the </u><u>similarity </u><u>theorem </u><u>:</u><u>-</u></h3>

  • If the ratio of length and breath of both the triangles are same then the given triangles are similar.

<u>Let's </u><u>Understand </u><u>the </u><u>above </u><u>theorem </u><u>:</u><u>-</u>

The dimensions of rectangle 1 are x and y

<u>Now</u><u>, </u>

  • Rectangle 2 is made by multiplying each dimensions of rectangle 1 by a factor of k .

Let assume the value of K be 5

<u>Therefore</u><u>, </u>

The dimensions of rectangle 2 are

\sf{ 5x \:and \:5y }

<u>Now</u><u>, </u><u> </u><u>The </u><u>ratios </u><u>of </u><u>dimensions </u><u>of </u><u>both </u><u>the </u><u>rectangle </u><u>:</u><u>-</u>

  • \bold{Rectangle 1 =  Rectangle 2}

\bold{\dfrac{ x }{y}}{\bold{ = }}{\bold{\dfrac{5x}{5y}}}

\bold{\blue{\dfrac{ x }{y}}}{\bold{\blue{ = }}}{\bold{\blue{\dfrac{x}{y}}}}

<u>From </u><u>above</u><u>, </u>

We can conclude that the ratios of both the rectangles are same

Hence , Both the rectangles are similar

<h3><u>Answer </u><u>2</u><u> </u><u>:</u><u>-</u><u> </u></h3>

<u>Here</u><u>, </u>

We have to proof that, the

  • Perimeter of rectangle 2 = k(perimeter of rectangle 1 )

In the previous questions, we have assume the value of k = 5

<h3><u>Therefore</u><u>, </u></h3>

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

Perimeter of rectangle 1

\sf{ = 2( length + Breath) }

\bold{\pink{= 2( x + y ) }}

Thus, The perimeter of rectangle 1

Perimeter of rectangle 2

\sf{ = 2( length + Breath) }

\sf{ = 2(5x + 5y) }

\sf{ =  2 × 5( x + y) }

\bold{\pink{= 10(x + y) }}

Thus, The perimeter of rectangle 2

<u>According </u><u>to </u><u>the </u><u>given </u><u>condition </u><u>:</u><u>-</u>

  • Perimeter of rectangle 2 = k( perimeter of rectangle 1 )

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 2(x + y) = 10(x + y)}

\bold{\pink{2x + 2y = 5(2x + 2y) }}

<u>From </u><u>Above</u><u>, </u>

We can conclude that the, Perimeter of rectangle 2 is 5 times of perimeter of rectangle 1 and we assume the value of k = 5.

Hence, The perimeter of rectangle 2 is k times of rectangle 1

<h3><u>Answer 3 :</u></h3>

<u>Here</u><u>, </u>

We have to proof that ,

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<u>That </u><u>is</u><u>, </u>

  • Area of rectangle 1 = k²( Area of rectangle)

<h3><u>Therefore</u><u>, </u></h3>

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

<u>Area </u><u>of </u><u>rectangle </u><u>1</u>

\sf{ = Length × Breath }

\sf{ = x × y }

\bold{\red{= xy }}

<u>Area </u><u>of </u><u>rectangle </u><u>2</u>

\sf{ = Length × Breath }

\sf{ = 5x × 5y }

\bold{\red{ = 25xy }}

<u>According </u><u>to </u><u>the </u><u>given </u><u>condition </u><u>:</u><u>-</u>

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\sf{ xy = 25xy }

\bold{\red{xy = (5)²xy }}

<u>From </u><u>Above</u><u>, </u>

We can conclude that the, Area of rectangle 2 is (5)² times of area of rectangle 1 and we have assumed the value of k = 5

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