Staying Rational in a World of Tweets, Fake News, Alternative Facts and Sound Bites

by William P. Hall
william-hall@bigpond.com
http://www.orgs-evolution-knowledge.net 

( based on a presentation at Vic Skeptics Café, 19 June 2017 at the Clyde Hotel, Carlton, Vic)

Today we are living in a world of Twitter, Facebook, Instagram, and blogs where anyone (even President Trump) can instantly post their ideas to the world for essentially no cost. We are also living in dangerous times where exploding human populations and technologies are affecting the planet’s climate and natural resources where extreme concentrations of wealth and power, warfare, epidemics, climate extremes, ecological collapses and famine threaten humanity’s survival. Unsurprisingly there are often conflicts between vested interests seeking wealth, power and control versus those concerned with the futures of our descendents and of humanity in general. Both are heavy users of the new media. View More Staying Rational in a World of Tweets, Fake News, Alternative Facts and Sound Bites

Dr Ken Harvey receives the AM

Prominent Vic Skeptic Dr Ken Harvey has been awarded the AM (Member of the Order of Australia) in the 2017 Queen’s Birthday Honours List.
Read all about it HERE

More on Dr Ken Harvey
https://skeptics.cafe/2015/09/13/dr-ken-harvey-on-the-radio-chiropractic-advertising/
https://skeptics.cafe/videos/dr-ken-harvey-chiropractors-abcrn-10915/
https://skeptics.cafe/2012/04/02/dr-ken-harvey-awarded-choice-consumer-champion/
https://skeptics.cafe/2012/03/06/tga-dr-ken-harvey-and-sensaslim-news/
https://skeptics.cafe/2011/06/11/the-sensaslim-libel-action-against-dr-ken-harvey-how-did-it-happen/
https://skeptics.cafe/2010/06/26/dr-ken-harvey/

Death of Eric Fiesley

We are saddened to report that Eric Fiesley died earlier this month, several weeks after a road accident in Mildura.

A retired TAFE teacher, Eric was well-known to the Skeptical community, as a good friend and as contributor to and administrator of on-line Skeptical and social chat groups, and an enthusiastic supporter of campaigns. We also remember his devotion to caring for his sick wife Pam over many years.

Eric played euphonium with the Mildura and District Brass Band. He was also an administrator of both the Ulysses Social Club (for motorcyclists over the age of 40) , and the Mildura Working Man’s Club.

He will be missed.

 

“Mixed Bag” Questions October 2017 – Answers

1. A small hotel or boarding house

2. Accept Morris Mini OR Mini Cooper

3. Bugsy

4. Ladybird/ladybug

5. To Kill a Mockingbird

6. Two (Freshwater and Saltwater)

7. Dame Judy Dench

8. The Goon Show

9. Rolls Royce Merlin

10. 10CC

HARDER:

11. Had its genome (DNA) mapped

12. Canoe / Kayak

13. Kenneth Williams

14. The Wise Men visiting the baby Jesus

15. D. They hanged him anyway

16. Burning

17. Sweden

18. Whitney Houston

19. Taff

20. Clouds

“Mixed Bag” Questions August 2017 – Answers

1. Two hours (108 minutes)

2. Moonies

3. Perspective

4. Sting a.k.a. Gordon Sumner

5. Over eyes, bridge of nose, etc (they are eyeglasses that sit on your nose)

6. False. He was actually once a criminal on the FBI Ten Most Wanted List.

7. Beauty and the Beast

8. Halley’s Comet

9. Marni Nixon

10. 50%

HARDER:

11. 1937

12. Timor Sea

13. Pride

14. Ogee

15. Toni Collette. The film was Spotswood

16. Little cords

17. Portmanteau words: (Two or more previously existing words run together)

18. Dusty Springfield

19. Guinea-Bissau, Nauru, Peru, Tuvalu, Palau, Vanuatu

20. Regina

“Mixed Bag” Questions July 2017 – Answers

1. Eat them (little cakes)

2. They go out in the midday sun

3. Bob Fosse

4. The Righteous Brothers

5. Glenn Close

6. Motor home, motorised caravan etc.

7. Brother Cadfael

8. London

9. A butt (barrel) of wine

10. Ripley

HARDER:

11. Gerald Ford

12. 1970s

13. Walk

14. 5 (Diana, Randolph, Sarah, Marigold, Mary)

15. 3

16. Lightning

17. 200

18. Tetrahedron, cube, octahedron, dodecahedron, icosahedron

19. Panama

20. John Paul I

August 2017 Logic & Maths Problems – Solutions

1. 18

2. Daniel is 7, Jessica is 4

3. 32
They change square colour when they move. So if they were all on the 32 white squares or the 32 black squares they could not attack each other.

4. 102 pages
To eat 300 pages with the books stacked as shown, the bookworm would have to eat from p100 Vol 1 to p1 vol 3

5. The red door
Process of elimination as follows:

6. a. 13; b. 18; c. 3

7. a

8. a. Keith b. first c. Adrian

9. 2/3 (not 1/2)

You know that you do not have Bag B (two black marbles) so there are three possibilities

You chose Bag A, first white marble. The other marble will be white
You chose Bag A, second white marble. The other marble will be white
You chose Bag C, the white marble. The other marble will be black

So 2 out of 3 possibilities are white.

Why not 1/2? You are selecting marbles, not bags.

10. 125 m3..

(half as high means all three dimensions are in the ratio 0.5.

The smaller pyramid’s volume is

0.5 × 0.5 × 0.5 = 0.125

 that of the larger pyramid.)

July 2017 Logic & Maths Problems – Solutions

1. 6

2. 12

3. 44 years.

Let the number of years elapsed on the lease be y.
Then the number of years left on the lease is (99-y).

3y / 4   =  3(99-y) / 5

15y = 12(99-y)

15y = 1188 – 12y

27y = 1188

y = 44

4. “You forgot the spaces between ‘Johnson’ and ‘and’ and ‘and’ and ‘Smith’ ”

5. 720 times

6. 2 cakes

7. 60 legs
For any centipede to have the largest possible number of legs means the largest possible share of the available 122 legs.
Obviously this implies only one centipede. That leaves 10 heads shared between spiders and ants. Since spiders have more legs than ants, we are looking for the least possible number of spiders (1) hence 9 ants.
1 spider and 9 ants account for (8 + 54) = 62 legs. That leaves 60 legs available to the centipede.

8. Carrots $2.99, spring onions $1.99, parsnips $5.99

9. A = 8 B =2 C=1

111A+111B+111C=110B+1001C which reduces to
111A+B=890C

The most 111A can be is 999 so the most (111A + B) can be is 999 + 8 = 1007.
So C can only be 1. For example, if C was 2, 890C would be 1780.
Therefore 111A+B=890

A must be 8 for the equality to be possible,

=> B=2

10. 1009

A possible way of approaching the problem is as follows:
a. Eliminate even numbers.
b. Eliminate numbers ending in 5.
c. Eliminate numbers using the “divisible by 3” rule. Any whole number which is divisible by 3 has digits which add up to a multiple of 3
d. Of the remaining possibilities (1001, 1003, 1007, 1009, 1013, 1019, ………..) begin with the lowest and attempt to divide by successive primes beginning with 7. You can stop when you get to a square root. As 31 squared is 961 and 37 squared is 1361, that effectively means if you get to 31 in your attempt to divide by successive primes you’ve gone far enough.
1001 = 7 X 143 = 7 X 11 X 13 NOT PRIME
1003 is not divisible by 7, 11, or 13. However it is divisible by 17.
1003 = 17 X 59 NOT PRIME
1007 is not divisible by 7, 11, 13 or 17. However it is divisible by 19
1007 = 19 X 53 NOT PRIME
1009 is not divisible by 7, 11, 13, 17, 19, 23, 29 or 31. It is therefore PRIME.