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Assuming that $300 and $50 are the respective cost(s) of the premium and discount tickets, if we didn't buy any premium tickets, we could afford to buy 12 discount tickets at $50 because 12($50) = $600

The number of total sets possible by choosing any 6 people from 15 = C(15,6) = 5005

But the only ones we're intereseted in are the ones where we choose 4 men from 8 and 2 women from 7 = C(8,4) * C(7,2)  = 1470

So.....1470/5005 = about 29.3%

First, convert both of them to improper fractions:

-4 1/5

1. Multiply the whole number (4) by the denominator (5) to get 20 (ignore - signs for now)

2. Add the numerator (1) to get 21

3. Put it over the denominator to get 21/5. Now add a minus sign to get $${\mathtt{\,-\,}}{\frac{{\mathtt{21}}}{{\mathtt{5}}}}$$

Do the same for the other fraction to get $${\mathtt{\,-\,}}{\frac{{\mathtt{10}}}{{\mathtt{3}}}}$$

Now find the Least Common Multiple (LCM) for the denominators, 3 and 5. Keep multiplying 5 until you get a number that works for both 3 and 5. In this case, it is 15. 3*5=15 and 5*3=15. Multiply them by those numbers so that the denominator is 15:

$${\mathtt{\,-\,}}{\frac{{\mathtt{21}}}{{\mathtt{5}}}}$$ X $${\frac{{\mathtt{3}}}{{\mathtt{3}}}}$$ = $${\mathtt{\,-\,}}{\frac{{\mathtt{63}}}{{\mathtt{15}}}}$$

Do the same for $${\mathtt{\,-\,}}{\frac{{\mathtt{10}}}{{\mathtt{3}}}}$$, but multiply by 5/5 instead of 3/3. This will give you $${\mathtt{\,-\,}}{\frac{{\mathtt{50}}}{{\mathtt{15}}}}$$. Now that they both have the same denominator, you can multiply them together.

$${\mathtt{\,-\,}}{\frac{{\mathtt{50}}}{{\mathtt{15}}}}$$ X $${\mathtt{\,-\,}}{\frac{{\mathtt{63}}}{{\mathtt{15}}}}$$

Cross reduce if you're in the mood, but to keep things short I'll skip it.

-50*-63=3150 (remember that negative*negative=positive)

Leave the denominators alone.

This gives you $${\frac{{\mathtt{3\,150}}}{{\mathtt{15}}}}$$. You can then simplify it. 3150 is divisible by 15, so you can divide to get $${\frac{{\mathtt{210}}}{{\mathtt{1}}}}$$, or just plain 210.

Good luck.

To find the size of the angle when you know the value of sin, use invsin or (2nd) sin:

If sin A = 0.36 then invsin(0.36) ≈ 21.1°

You can think of negative numbers as cold numbers and positive numbers as hot

so if you add hot water the bath gets hotter    ++=+

the same as if you take away cold air from a room the room gets hotter   - - = +

-19.6 - 12  =  -19.6 + -12  =  -31.6

A negative exponent indicates that the number is in the wrong location (top to bottom):

1) If the number is in the numerator with a negative exponent, move the number to the denominator and make     the exponent positive.

    Examples:  6^-1  becomes  1 / 6

                     5^-3  becomes  1 / 5^3

                     a^-5 · b^4 · c^-8  becomes  b^4 / ( a^5 · c^8 )

2) If the number is in the denominator with a negative exponent, move the number to the numerator and make     the exponent positive.

    Examples:  1 / 7^-3  becomes  7^3

                     6 / a^-2  becomes  6 · a^2

Difficult example:

                     ( a^-4 · b^5 · c^-2 ) / ( a^3 · b ^-4 · c^-8 )  

                             becomes  ( b^5 · b ^4 · c^8 ) / ( a^4 · a^3  · c^2 )  =  ( b^9 · c^6) / a^7

cot( π/2 - y ) = tan (y )

sin²(y) - tan²(y)

This is:

sin²(y) - [ sin²(y) / cos²(y) ]   =   [ cos²(y) · sin²(y) / cos²(y) ] - [ sin²(y) / cos²(y) ] 

      =  [  cos²(y) · sin²(y) - sin²(y) ] / cos²(y)  

      =  [ sin²(y) · ( cos²(y) - 1 ) ] / cos²(y) 

      =  [ sin²(y) · ( - sin²(y) ) ] / cos²(y) 

      =  - sin^4(y) / cos²(y) 

Does this help?  

If a potato has an odd number of eyes, does that mean there is a half a cat somewhere.

Which half would that be Gino?

8 cookies is the answer

Hi Linlee,

I just want to complement you on your wonderful manners.  

Thank you we all really appreciate keen students with good manners.    

I like your comprehensive answer Geno   

-3y - 6(3z - 3y) - 5z        Use the distributive property: multiply -6 times 3z and -6 times -3y

-3y - 18z + 18y - 5z        Combine the y's and combine the z's

15y - 23z

For every real number x, x² is positive; thus, the absolute value of x² is x².

If s = success and f = failure, the possibilities are:

s = 2 and f = 7     P = 9C2 x (5/7)^2 x (2/7)^7

s = 3 and f = 6     P = 9C3 x (5/7)^3 x (2/7)^6

s = 4 and f = 5     P = 9C4 x (5/7)^4 x (2/7)^5

s = 5 and f = 4     P = 9C5 x (5/7)^5 x (2/7)^4

s = 6 and f = 3     P = 9C6 x (5/7)^6 x (2/7)^3

s = 7 and f = 2     P = 9C7 x (5/7)^7 x (2/7)^2

Find each of the above probabilities and add them together for the total probability.

10^-1  =  1/10

30 x 10^-1  =  30 x 1/10  =  30/10  *  3

17/4 = 4.25

Since there is an increase (1280 more) use a + sign:  + 1280

the answer is 21

Thanks for that website reference, Melody.....I can never quite summon the words to explain how to do this - even though it's easy !!! As they say....the pics on the website are worth 1000 words !!! (Well....I don't actually know if anyone used those exact words.....)

The quick way to do this:  you need to write the number with the decimal point after the first non-zero digit

(that is as 2.24 x 10^? ) with the exponent on 10 the number of places that you have to move it to get there.

Since it starts at the far right of the number and moves 9 places, the exponent will be 9.

Answer:  2.24 x 10^9

S = the semi-perimeter = (15 + 12+ 10) / 2  = 18.5

So we have

A = √[(18.5)*(18.5-15)*(18.5-12)*(18.5-10)]= √[18.5 * 3.5 * 6.5 * 8.5]= 59.8 m²

BTW......this "formula" is known as "Hero's Formula" or "Heron's Formula".....however, many mathematicians believe that it bears the distinct "pawprint" of Archimedes...the proof of it, geometrically, is somewhat involved......something at which Archimedes excelled !!!

Try looking through this.

Since  a = 15  b = 12  and  c = 10

then  s  =  ( 15 + 12 + 10 ) / 2         --->         s  =  37/2  =  18.5

Area  =  √( 18.5 x ( 18.5 - 15 ) x ( 18.5 - 12 ) x ( 18.5 - 10 ) )     =  √ ( 18 x 3.5 x 6.5 x 8.5 )

         =  √ 3480.75  ≈  59.0

This formula is known as Hero's Formula or Heron's Formula.

I don't understand what you mean isi?

I am not sure what anon wants either.

Maybe

$$2^{40}=2^{10}*2^{10}*2^{10}*2^{10}$$

$${{\mathtt{2}}}^{{\mathtt{10}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{10}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{10}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{10}}} = {\mathtt{1\,099\,511\,627\,776}}$$