Functional Analysis MIT question/problem

Q: Consider a real-valued function f : R −→ R which is locally integrable in the

sense that

(5.2) gL(x) =

f(x) x ∈ [−L,L]

0 x ∈ R \ [−L,L]

is Lebesgue integrable of each L ∈ N.

A: 

f: R ---> R

locally integrable

(5.2) Greenland[gL](x) =

function (x) x is an element of [-L,L]

0 x is an element of R divided by [-L,L]

Lebesgue integrable = L is an element of  N

5.2gL f (R) =

 L = R + L + (- L) 

A little confused on this.

Hope you can correct and leave a comment.

Thanks.

Set Theory -Calculus Question.

Write carefully using English and using it correctly.

Attempt to answer all the questions. You may use the previous questions in your answers.

You have 2 hours and 30 minutes.

1. Show that if q Î Q is a square in Q, then 2q is not a square in Q.

q equals q 

  =    

# Q #  

 =


 =   =
#Q# #Q# = 2Q
 =   =


The above is not a square in Q rather a set of squares in Q. =).


Thanks.


Think this is incorrect?


Please comment.

Trigonometry Question.

Q: 

Sin B = L of opposite side

              --------------------

              L of Hypotenuse

Tan B = L of opposite side

                ----------------------

               L of Adjacent side

Sin B = 50/14 - Approx - 3.6

Tan B = 50/48 - Approx - 1.02


Sin = 50/14
50 = 14 * sin = 3.6
Sine = 3.6
+ on going below....

Sin B = 14,28,42,49 {3.5},50 3.6

Sin B = 24/25 - Approx - 0.95

Tan B = 24/7 - Approx - 3.4

Sin B = 7/25 - Approx - 3.6

Tan B = 7/24 - Approx - 3.4

Sin B = 7..24 - Approx - 3.4

Tan B = 7/245 - Approx - 3.6

Think this is incorrect?

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{Decision Making} - CVP Analysis - Capacity Usage of a Leisure Centre

Advanced -


A local government authority owns and operates a leisure centre with numerous sporting facilities, residential accommodation, a cafeteria and a sports shop. The summer season lasts 20 weeks including a peak period of six weeks corresponding to the school holidays. The following budgets have been prepared for the next summer season:


Accommodation 
60 single rooms let on a daily basis
35 double rooms let on a daily basis at 160 per cent of the single room rate.
Fixed costs £29,900
Variable costs £4 per single room per day and £6.40 per double room per day.
Sports Centre 
Residential guests each pay £2 per day and casual visitors £3 per day for the use of facilities.
Fixed costs £15,500
Sports Shop
Estimated contribution £1 per person per day.
Fixed costs £8250
Cafeteria 
Estimated contribution £1.50 per day.
Fixed costs £12,750


During the summer season the centre is open seven days a week and the following activity levels are anticipated.


Double rooms fully booked for the whole season.


Single rooms fully booked for the peak period but at only $0 per cent of capacity during the rest of the season.
30 casual visitors per day on average.


Questions -
a): Calculate the charges for the single and double bedrooms assuming that the authority wishes to take £10,000 profit on accommodation.
b): Calculate the anticipated total profit for the leisure centre as a whole for the season.


Ans -
a): £4 * 60 = £240 
£6.40 * 35 = £224 (approx)


£240 * 30 * 7 = £7200 {*7} = £50400  {1 week}
£224 * 35 * 7 = £7800 {*7} = £62400 {1 week}


Total for 1 week = £112,800 
Minus £10,000 profit = £102,800


Equals = £102,800 


£112,800 * 20 = £2,256,000 {Total Profit}
Fixed Costs = £29,900 
  £2,256,600
- £   29,900
=  2,226,700
-     10,000
=  2,216,700
Total Charges : 39,900
Total Profit: 2,216,700


b): RVisitors: £2 * 7 = £14 
CVisitors: £3 * 7 = £21


£14 * 30 = £420 
£21 * 30 = £630


£420 * 30 = 12600
£630 * 30 = 19200


Total: 12,600 + 19,200 = 31,800
£10,000 - Accommodation Authority Fee
£15,500 - Fixed Costs
= £6,300 {Total Profit Remaining}


Thank you very much =)

Clinical Professional Decision Making Question

3. Construct a five-year leadership plan for yourself, then in one-page to two-page paper, outline a development plan based on your ultimate health leadership position goal. How long will it take you to read your ultimate goal?

Answer: 

Year 1: Reduce Meds - 5ml - 23

Year 2: Reduce Meds - 5ml - 24

Year 3: Reduce Meds II - 50mg Jan - 25

Year 3: Reduce Meds II - 50mg Dec - 25

Year 4: Reduce Meds II - 50 mg Jan - 26

Year 5: Meds II - 50mg March - 27

Year 5: Meds II - Steady - 27+  

This was an example, particularly difficult, but enabled to profess the correct analysis required for Clinical Management of the medication.

Think this is incorrect or have a solution?

Let me know....

Thanks,

Mohammad.

Probability Distributions - A Level Math

1. Two fair dice are thrown. If the scores are unequal, the larger of the two scores is recorded. If the scores are equal then that score is recorded. Let X denote the number recorded.

a) show that P(X = 2) = 1/12 and draw up a table showing the probability distribution of X

Ans:

Two fair dice {6 sided}

Scores - unequal:

1. 1+1=1 {X =1}
2. 2+1=3 {X = 2}
3. 1+2=3 {X = 2}
4. 2+2=4 {X = 2]
5. 2+2=4 {X = 2}
6. 2+3=5 {X = 3}
7. 3+2=5 {X = 3}
8. 3+3=6 {X = 3}
9. 3+3=6 {X = 3}
10. 3+4=7  {X = 4}
11. {X = 4}
12. {X = 4}
13. {X = 4}
14. {X = 5}
15. {X = 5}
16. {X = 5}
17. {X = 5}
18. {X = 6}
19. {X = 6}
20. {X = 6}
21. {X = 6}

etcetera...

P(X=2)

 X=2 == 4/21 if 1/12 then {1.9/21} e.g. 

Calculation Based On: 2/21 (Equal)
0.95/21 = 0.045[recurring]

Calculation Based On: 2/21 (Unequal)

0.95/21 = 0.045[recurring]

Thank you.

Think this is incorrect??

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Probability - Dice

A pair of dice is rolled. What is the probability of getting a sum of 2?     

None of the above.

 

Ans: 

 

{2 Dice}

1. 1+1 = 2 
2. 1+1=2
3. 1+2=3
4. 2+1=3
5. 2+2=4
6. 2+2=4
7. 2+3=5
8. 3+2=5
9. 3+3=6
10. 3+3=6
11. 3+4=7
12. 4+3=7
13. 4+4=8
14. 4+4=8
15. 4+5=9
16. 5+4=9
17. 5+5=10
18. 5+5=10
19. 6+5=11
20. 5+6=11
21. 6+6=12
22. 6+6=12 
     
    2/22 = 1/11
    
    
    Ans: None of the above.
    
    Think this is incorrect???
    
    
    Please leave a comment.
    Thanks.

Equatorial Belt.

4. Equatorial belt

A snug-fitting belt is placed around the Earth's equator.  Suppose you added an extra 1 meter of length to the belt, held it at a point, and lifted until all the slack was gone.  How high above the Earth's surface would you then be?  That is, find h in the diagram below.

Assume that the Earth is a perfect sphere of radius 6400 km, and that the belt material does not stretch.  An approximate solution is acceptable.

A:

For this calculation we need PiR^2

As we know 1 meter is 'h' we calculate 'r' by adding 'h' into 'r', basically we calculate, how many 'h' go into 'r' and add 1. So, I would approximate 6, so the radius is 6400, so we do the following sum.

6+1 (7)*3.14{pi for arguments sake}^2 =

2*3.14 = 6.28
2*6.28 = 12.56 {4}
2*12.56 = 25.12 {8}
25.12 - 3.14 = 21.98 {7}

21.98 * 21.98 =
Approx 22*22 = 484

Precise 0.2 * 21.98 =
or 21.98/5 = 4.396
8.792
13.188
17.474
21.98

484 - 4.396 = 479.604 {Precise Answer}

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Folded Sheet of Paper

Q: 1. Folded sheet of paper 

A rectangular sheet of paper is folded so that two diagonally opposite corners come together.  If the crease formed is the same length as the longer side of the sheet, what is the ratio of the longer side of the sheet to the shorter side?

A {probably =) } :

Longer Side: 4:3

Shorter Side: 3:4

(First number being higher than the second in the ratio).

 Why??

 This is because the shorter side if tilted to the longer side on the right as you see, then it is only 3/4 of the size, this is represented in ratio.

Think this is incorrect expirementation???

Leave a comment, pleased to hear what you think, =).