geno3141

a + b > c     or     a + c > b     or     b + c > a

You are choosing 3 out of 24 students = ₂₄C₃.

Either 2 of the 14 sociology students = ₁₄C₂ and 1 of the 10- non-sociology students = ₁₀C₁

or       3 of the 14 sociology students = ₁₄C₃ and none of the 10- non-sociology students = ₁₀C₀.

[ ₁₄C₂ x ₁₀C₁  +  ₁₄C₃ x ₁₀C₀ ] / ₂₄C₃

=  [ 91 x 10  +  364 x 1 ] / 2024 

=  [ 910 + 364 ] / 2024

=  1274 / 2024

cos^-1( -2·sqrt(2) )  does not exist.

Let's say that Zach spent  Z  on his lunch.

Brain (Brian?) spent $4.00 lest than 3 times Zach:  Brain spent  3·Z - 4.00.

Together, they spent  $25.00:     Z + (3Z - 4.00)  =  25.00

                                                            4Z - 4.00  =  25.00

                                                                       4Z  =  29.00

                                                                         Z  =  $7.25

                          3·Z - 4.00   --->   3(7.25) - 4.00  =  $17.75          

The custormer has a choice of 6 types of crust.

The custormer can choose 2 meats from 4 meats:  ₄C₂  =  6.

The custormer can choose 2 vegetables from 9 vegetables:  ₉C₂  =  36.

Since these are independent choices, mutiply these 3 answers together:  6 x 6 x 36  =  your answer.

To find where they intersect, set the two equations equal to each other:  x² + a  =  ax  and solve:

--->      x² + a  =  ax      --->     x² - ax + a  =  0

Using the quadratic formula:  x  =  [ a +/- sqrt( (-a)² - 4·1·a } / ( 2·1)    

--->     x  =  [ a +/- sqrt( a² - 4a ) ] / 2

If they intersect in only one point,  a² - 4a  must equal zero.

If     a² - 4a  =  0     --->     a(a - 4)  =  0

either     a = 0    or    a = 4

The sum of these two possibilities is  0 + 4  =  4

There are more irrational numbers than there are rational numbers.

If the odd numbers of the first die are painted red, the numbers 1, 3, and 5 are painted red.

On the second die, the only number that is painted red is the number 3.

To get a red numer to be rolled on the first die has a probability of 3/6.

To get a red number to be rolled on the second die has a probability of 1/6.

To get a red number to be rolled on both dice has a probability of (3/6) x (1/6)  =  3/36  =  1/12

-2u³y              25x³

-------     -     ---------      =     

15xz⁵          14u²y²

To subtract fractions, you need a common denominator; in this case, the common denominator is found by multiplying both denominators together:

     Denominator of each fraction =  15xz⁵ · 14u²y²  =  210xz⁵u²y²  

So, both the numerator and denominator of the first fraction must be multiplied by 14u²y²

     Numerator of the first fraction =  -2u³y · 14u²y²  =   -28u⁵y³

And, both the numerator and denominator of the second fraction must be multiplied by 15xz⁵

     Numerator of the second fraction  =  25x³ · 15xz⁵  =  375x⁴z⁵

The problem is now:     [ -28u⁵y³ / 210xz⁵u²y²  ]  -  [ 375x⁴z⁵ / 210xz⁵u²y²  ]

Combining the two numerators:  -28u⁵y³ - 375x⁴z⁵  over the common denominator:  210xz⁵u²y²  gives you your answer"

     =     ( -28u⁵y³ - 375x⁴z⁵ ) / 210xz⁵u²y² 

Probability that it will rain:

-- on both Saturday and Sunday  :  0.46 x 0.54  =  0.2484

-- on Saturday but not on Sunday:  0.46 x 0.46  =  0.2116

-- on Sunday but not on Saturday:  0.54 x 0.54  = 0.2916

On Saturday or Sunday (inclusive or  --  possibly both)  =  0.2484 + 0.2116 + 0.2916  =  0.7516

On Saturday or Sunday (exclusive or  -- one or the other, but not both)  =  0.2116 + 0.2916  =  0.5032

This may be somewhat confusing because "not Sunday"  =  1.00 - 0.54  =  0.46 (the same as "on Saturday") and

"not Saturday"  =  1.00 - 0.46  =  0.54 (the same as "on Sunday").