Selecting A Dog Bed by Kevin Gawricki
Some general guidelines for selecting a bed for your dog.
Crate Trained Dog
If you are crate training a pup use a soft blanket or towels until pup is ready for a bed.
A crate pad is the best way to go. It offers a plush bumper all around the bed which provides cushioning from the crate.
Always remember to measure your crate to assure proper fit.
Machine washable is also a good feature that makes keeping your pet’s bed fresh and clean easier.
Senior Dogs
Some senior dogs will prefer the orthopedic dog beds. They are made from high density foam that will help to cushion sore joints and muscles.
Easy to Clean
Beds that are machine washable offer a easy way to keep your pet’s bed fresh and clean.
Beds that are not machine washable should offer at least a removable cover that can be easily washed.
Consider How Your Pet Likes to Sleep
If your pet likes to stretch out , than consider a longer bed.
Does your pet like to rest his/her head on something. Than consider a bed with some type of bumper.
Does your pet like to curl up while he/she sleeps. Than consider an oval, round, or snuggle bed.
Does your pet need to be above the floor, consider a raised bed.
With these general guidelines and a little thought you are sure to get a bed that will be a perfect fit for your pet.
Kevin is owner of Dog Gone Good Stuff which he founded because of the need for a place that caters specifically to dogs and their owners specific needs. Please feel free to visit Dog Gone Good Stuff for all your dogs needs. Web address www.doggonegoodstuff.com
Kevin Gawricki
http://www.articlesbase.com/pets-articles/selecting-a-dog-bed-62284.html
What is the probability of selecting an ace or a spade from a deck of 52 cards?
What is the probability of selecting an ace or a spade from a deck of 52 cards?
uhhh 1 in 52 ….. duh
1.92307692%
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Ah, 1 in 52 chance
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1 in 2 as long as you don’t include the joker
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Pretty high if one is an ace, one is a spade and the other 48 are oversized greeting cards. Have a nice day
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It depends on how much money you stand to lose if you need one.
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There are 4 aces so 4/52 =1/13
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There are 13 spades in a deck and 4 aces. One of the aces is a spade which gives you 16 cards out of your 52 cards.
16/52 reduces to 8/26 then to 4/13.
So the probability is 4/13
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12 (2-king) + 4 aces = 16
16 over 52 is 4 over 13
=4/13 chance
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There are 13 spades plus 3 other aces = 16
16/52
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P(Ace) is 4/52 = 1/13
P(Spade) is 13/52 = 1/4
P(Ace and Space) = 1/52
therefore P(Ace or Spade) = 1/13 +1/4 – 1/52 = 4/13
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You’ve had some pretty ignorant answers to this one! I agree with Sara S & Tasty who have supplied the correct answer.
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There are 52 cards in total
13 cards are Spades.
4 cards are aces.
1 card is both.
So, the probability of selecting an ace OR a spade is (13 + 4 -1)/52
= 16/52
= 4/13
= 30.7692%
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I got 100% on my Mathods SAC
selecting an ace from a deck of cards is 4/52 then you have to add the chance to selecting a spade from the deck 13/52. then you need to take away the 1/52 because the ace of spades is counted twice
I’m not good with fractions but google can give a decimal answer
(4/52)+(13/52)-(1/52)= 0.307692308
if someone answered 1/52… they are stupid
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good woman sara s, sense at last. of course there was something to think about and work out in the problem. probability is like that. work out the possible number of different events (spades + aces with sneaky ace of spades) and put it over the number of choices (eg 52 cards). that’ll do for a best answer, lass
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1 in 3.25
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4 Aces+12 spades= 16/52
4/52 + 1/4=17/52
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P(Ace) = 4/52, P(Spade) = 13/52, P(Ace v Spade) = P(A) + P(S) – P(AnS) = 1/13 + 1/4 – 1/52 = 4/13
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= 100%([4 + 4]/52)
= 100%(8/52)
= 100%(2/13)
= 200/13% or 15.38 6/13%
Answer: 2/13 or 15.38 6/13% or 2:11 ratio
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You have 13 cards that are spades. Each of these cards meets the requirement. There are three other aces in the deck that will meet the requirment. So, you have 13 + 3 cards that are winners.
The probability is then 16/52 = 8/26 = 4/13.
You have a 4 out of 13 chance of being successful, or in other words a probability of 0.30769230769230769230769230769231.
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