100 students were asked to ll out a form with three survey questions, as follows: H: Honor Roll C: Club membership (Robotics Club or Gaming Club)
D: Double-major
Survey results were as follows:
28 checked H (possibly non-exclusively), 26 checked C (possibly non-exclusively), 14 checked D (possibly non-exclusively)
8 checked H and C (possibly. non-exclusively), 4 checked H and D (possibly. non- exclusively), 3 checked C and D (possibly. non-exclusively)
And 2 checked all three statements.
1. How many students didn't check any of the boxes?
2. How many students checked exactly two boxes?
3. How many students checked at LEAST two boxes?
4. How many students checked the Clubs box only? [d]

1. 45

2. 9

3. 11

4. 17

Step-by-step explanation:

The computation is shown below:

Given  that

28 checked H

26 checked C

14 checked D

8 checked H and C

4 checked H and D

3 checked C and D

2 checked all.

Based on the above information

N(H)=28

N(C) 26

N(D) 14

N(H∪C) = 8

N(H∪D) = 4

We know that

= 28 + 26 + 14 - 8 - 4 - 3 + 2

= 55

1. The students who didnt check any kind of the box is

= 100 - 55

= 45

2) Students who checked exactly two boxes is

= 8 + 4 + 3 - 6

= 9

3) Students who checked a minimum of two boxes

= 8 + 4 + 3 - 4

= 11

4) Given N(C) = 26

Now we have to deduct N(C∪D) and N(H∪C) as it could be checked aside from the club

= 26 - 8 - 3

= 15

Here the could be deducted twice under both categories so it would be added one time to neutralize

= 15 + 2

= 17

.

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hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

cosine

step-by-step explanation: