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a) \(p = \dfrac{8}{32}\dfrac{6}{31} = \dfrac{3}{62}\)

b) \(p = \dfrac{6}{32}\dfrac{6}{31}= \dfrac{36}{992} = \dfrac{9}{248}\)

c) \(p=\dfrac{6}{32}\dfrac{5}{31} = \dfrac{15}{496}\)

so we need the integers, -4 < n < 2, i.e.

-3, -2, -1, 0, 1

Thanks 

Heuristics. That is a new word for me. Thanks.

Thanks to Rom and EP too  :)

\(P[\text{random student is female}] = \dfrac{22}{22+18} = \dfrac{22}{40}=\dfrac{11}{20}\)

\(P[\text{random student is Mathematics major}] = \dfrac{18}{40} = \dfrac{9}{20}\)

\(P[\text{getting scholarship A}] = 0.05 \\ P[\text{getting scholarship B}] = 0.01\\ E[\text{Value of scholarship A }]= 0.05(2000-10) + 0.95(-10) = $90\)

similarly

\(E[\text{Value of scholarship B}] = 0.01(4000-25) + 0.99(-25) = \$15\)

Thanks 

72 half gallons /( 1 2/3 cup/cake *.125 half gal/cup )   = 345.6   = 345 cakes  

Rom there might be something wrong with your display.

The question was “Does 62 look like a fraction?”

It wasn’t “Can you make 62 look like a fraction?”

I don’t know what the funny-looking Z, sideways U followed by a funny-looking Q means. Is that formula used to turn whole numbers into fractions? If I learn this, will whole numbers start looking like fractions to me?

Maybe the question asker knows that formula.

\(x+2 = \sqrt{3x+10} \\ (x+2)^2 = 3x + 10 \\ x^2 + 4x+4 = 3x + 10 \\ x^2 + x -6 = 0 \\ (x+3)(x-2) = 0 \\ x = -3, 2 \\ \text{however }\dots \text{when we plug these back in to check the solutions} \)

\(-3+2 \overset{?}{=} \sqrt{3(-3) + 10} \\ -1 \overset{?}{=} \sqrt{\text{of anything}} \\ \text{over the real numbers... no} \\ \\ 2+2 \overset{?}{=} \sqrt{3(2) + 10} \\ 4 \overset{?}{=} \sqrt{16} \\ \text{yes}\)

\(\text{So we see that x=-3 is an extraneous solution}\\ \text{arising because we squared things in order to solve the original equation} \\ \\ \text{the range of the square root function is only non-negative reals} \\ \text{and thus we need to be careful and check solutions whenever we square things to solve an equation}\)

\(62 = \dfrac {62}{1} = \dfrac{124}{2} = \dots \\ \mathbb{Z} \subset \mathbb{Q} \\ \text{so yes 62 is a "fraction", more accurately a rational number}\\ \text{it also happens to be a completely wrong answer in this context but that's another story}\)

1/3 means that 1 out of 3 coins is a quarter      1/3 x 288 = 96 Qs

1/6                                                                      1/6 x 288 = 48 Qs

See how that works?   Can you do the others?

Does 62 look like a fraction?

Assuming you can use the numbers more than one time....up to three times:

33 x 33 x 33 = 35937 possible 'outcomes'

any probability is between 0 and 1 so 62 is a non-starter

there are 32 total marbles in the bag

a) \(P[\text{drawing a green marble}] = \dfrac{6}{32}=\dfrac{3}{16}\)

b) \(P[\text{drawing a red marble}] = \dfrac{10}{32}=\dfrac{5}{16}\)

c) The easiest way to do this is find the probability of drawing a yellow marble and subtracting that from 1

\(P[\text{drawing a !yellow marble}] = 1-P[\text{drawing a yellow marble}] = 1 - \dfrac{12}{32} = \dfrac{20}{32} = \dfrac{5}{8}\)

There are no integer solutions for (a) and (b) !!

There is an integer solution for (c)! x=7 is a solution.

It is called "heuristics" in Math !!

Where is the triangle "below" ??!!

Look at the 5th question below yours for the answer !!!

no good answer... must have been late.

1.20X =0.9Y
 X/Y = 1.20 / 0.90
X/Y = 4/3

Equalateral triangle angles ALL equal 60 degrees

so    60 = 3x+27   solve for x = 11

        60 = 2y-4                     y= 32

then x+y = 11 + 32 = 43

Hey Rom.....    Can you explain why 'clearly p = 3 '     ?     I know you found the correct answer, but why couldn't you assume at that point p = 8?  or  p= 13    etc?      Thanx     ~EP

My best guess: 2 + 97 + 101 =200.     2  x  97  x  101 =19,594

Here is my 'Long Division' answer: