Math destroyer T3 #15

[Josh779]

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It is found that 14% of a population has blue eyes, while 25% are left handed, and 8% are both blue-eyed and left handed. What is the probability that a person randomly chosen from the population is neither blue-eyed or left handed?

A. 11%
B. 29%
C. 22%
D. 33%
E. 31%

I made a ven diagram and tried: 8-19=6 and 8-25=17 and would think to add those together to get the % of blue or left handed...but 23 is not an option.

Solution says: 14+25-8=31%
Why add then subtract instead of subtract then add?

 

[Josh779]

Another:
Simplify: SQ[8xy^2/(6x^3y)]

I'm getting the x in the numerator because once I subtract them I get x^-2 in the denom. so i bring it up to the numerator. The solutions says as an intermediate step that the x^2 is in the denominator. This is mainly the step that is throwing me off. Hope my question makes sense.

 

[LetsGo2DSchool]

It is found that 14% of a population has blue eyes, while 25% are left handed, and 8% are both blue-eyed and left handed. What is the probability that a person randomly chosen from the population is neither blue-eyed or left handed?

A. 11%
B. 29%
C. 22%
D. 33%
E. 31%

I made a ven diagram and tried: 8-19=6 and 8-25=17 and would think to add those together to get the % of blue or left handed...but 23 is not an option.

Solution says: 14+25-8=31%
Why add then subtract instead of subtract then add?

You have to add all the total possibilities first and then subtract the double counting.

Thus 14 + 25 will include every mofo with blue "and/or" left handed and then subtract the 8 which are double counted ones.

In the below diagram, let's say the red circle is the people with blue eyes intersecting with the green circle which is left handers. Thus blue eyes are the red circle + the black. Left handers are the green circle + the black. The black is the double counting.

So to correctly solve this you add up all the blue eyed people in the red (which includes black) + all the left handers in the green (which includes black too) and then subtract the double counting (black).

View attachment venn-diagram.png

 

[Josh779]

You have to add all the total possibilities first and then subtract the double counting.

Thus 14 + 25 will include every mofo with blue "and/or" left handed and then subtract the 8 which are double counting.

ohh since your double counting both blue and left handed individuals you have to subtract from the total blue and left handed individuals. Got it, thanks.

 

[rockclock]

Another:
Simplify: SQ[8xy^2/(6x^3y)]

I'm getting the x in the numerator because once I subtract them I get x^-2 in the denom. so i bring it up to the numerator. The solutions says as an intermediate step that the x^2 is in the denominator. This is mainly the step that is throwing me off. Hope my question makes sense.

SQRT[ (8xyy) / (6xxxy) ]

Cancel one x and one y:

SQRT[ (8y) / (6xx) ]

You can pretty much do whatever you want from there...the x^2 is the only part that gets a little cleaner since that's the only perfect square.

 

[LetsGo2DSchool]

Another:
Simplify: SQ[8xy^2/(6x^3y)]

I'm getting the x in the numerator because once I subtract them I get x^-2 in the denom. so i bring it up to the numerator. The solutions says as an intermediate step that the x^2 is in the denominator. This is mainly the step that is throwing me off. Hope my question makes sense.

Whether you keep the x in the numerator or denominator doesn't make a difference as long as the sign in front of the exponent is correct.

Thus,

x^2 in numerator = x^-2 in denominator

Similarly,

x^-2 in numerator = x^2 in denominator

The (-) in front of the exponent is just saying that it's the reciprocal (or inverse).

So in this problem, after you simplify both x's in the numerator/denominator, you are left with x^-2 (in numerator) or x^2 (in denominator).

 

[Josh779]

SQRT[ (8xyy) / (6xxxy) ]

Cancel one x and one y:

SQRT[ (8y) / (6xx) ]

You can pretty much do whatever you want from there...the x^2 is the only part that gets a little cleaner since that's the only perfect square.

I thought 8x-xxx would leave 8/-xx....thus 8xx

 

[Josh779]

ohhh I get it! I was confusing myself. Thanks for the help!

 

[rockclock]

I thought 8x-xxx would leave 8/-xx....thus 8xx

lol...I honestly have no clue what you're trying to say bud.

My point was just that if wrapping your head around the rules for subtracting exponents is troublesome, you can always write it out...i.e. x^3 as xxx

After that you can cancel one-for-one in the numerator and denominator just like when you first learned algebra back in the day...exponents are just a shorthand, aint no thing haha

 

[Josh779]

haha ya I don't know. Never really had trouble with it before. Maybe it's time I focus on a different subject for the day.