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Its the study of topics such as quantity,structure,space and change.

No i mean to round to the nearest hundreth, i meant the numbers

A bag contains 2 white b***s,3 black b***s and 4 red b***s. In how many ways can 3 b***s be drawn form the bag, if at least one black ball is to be included?

I think this question is just about dead.

If three b***s are chosen at random then

P(at least one is black)

= 1-P(none are black)

$$\\=1- \frac{6C3}{9C3}\\
=1- \frac{20}{84}\\
=\frac{16}{21}\\$$

This is the same as Alan got and the same as DavidQD and CPhill would have gotif carried to its natural conclusions

We did not do it all the same way but this part of our answers are all the same.

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However I do not think that if is correct to say that there are 84 combinations of these 3 b***s or that there are 20 combinations where none are black.  BECAUSE some of these combinations are indistinguishable from each other.

There are only 6 possible outcomes if order does not count.  I think it is correct to call these outcomes combinations  

and there are 19 possible outcomes if order does count.  I think it is correct to call these outcomes permutations 

let x = 69,

so 2 x 69 is 138.

138*138 (Squaring) = 19044

$${\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{69}}\right)}^{{\mathtt{2}}} = {\mathtt{19\,044}}$$

For part a) I think there are only 15 sets of 3 pairs.  If the players are numbered 1 to 6 then the sets are as follows:

.

$$97\% \;of\; 1 = 97\% \;of \; \frac{100}{100}= \frac{97}{100}\;\;or\;\;0.97$$

(-5)(-3)-3 = (-5)*(-3)-3 

multiplication before subtraction

-5*-3=+15

15-3=12

Generally speaking the highest power(degree) tells you how many directions the graph will have

sometimes it can be deceptive because sometimes the graph tries to turn around but does not quite make it.

I am using VERY lay terms here.  

Lets look at some graphs.   a,b,c,...are just constants     LET a>0       

Now I am going to say what I think the degree of each of these will be  (I'll do the top row)

1)  It has five directions so the degree is probably 5 (it has to be odd greater than or equal to 5)

It ends on the right where y is positive so the leading coefficient is positive  

the constant (without an x)  is the y intercept

It is probably    $$y=ax^5+bx^4+cx^3+dx^2+ex-2$$

2)  4 directions ends down so probably     $$y=-ax^4+bx^3+cx^2+dx+0$$

3)  this one has got a kink - that is where it has tried to turn around but has not quite made it.

the degree is most likely 7                      $$y=-ax^7+bx^6+cx^5+dx^4+ex^3+fx^2+gx-2$$

4)  3 directions     $$y=-ax^3+bx^2+cx+3$$

Maybe you can try the others    

Maybe this will help?

1+1=window

Do you see it? It's called picto-math.

Thanks Geno,  

If the probability of spinning an A on spinner 1 is 3/4 and the probability of spinning E on spinner 2 is 1/3, what is the probability of spinning at least one vowel?

Geno's answer is really good and will work perfectly.

Here is another method  

first I will assume that the only vowel on 1 is A so P(vowel on 1) = 3/4

and  the only vowel on 2 is E    so  P(cowel on 2) = 1/3            (Geno made the same assumption)

I could add those together but if I did I would be counting the probability of getting 2 vowels twice

P(2 vowels) = 3/4*1/3 = 1/4

so

P(of spining at least one vowel ) =      $$\frac{3}{4}+\frac{1}{3}-\frac{1}{4}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$

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$$\boxed{P(A or B) = P(A)+P(B)-P(AandB)}$$

This is the 2 spinner outcomes.  So how if you just add them up the cross over bit will be counted twice?

Thanks Geno   

just divide both sides by 8  

m+m+2m

A mat + another mat + 2 more mats = 1+1+2=4 mats

When you are adding or subtracting with letters is is easier to think of  them as things.  :)

m+m+2m = 1m+1m+2m = 4m

a)  Choose 2 from 6 to be the first pair that is 6C2 =15

then choose 2 from 4 for the second pair   4C2 = 6

So I think that the answer for part a is      6C2*4C2 

$${\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{90}}$$

There are 90 pair combinations.      

Now I don't understand the question for part b.  Do the 4 new ones partner each other of do they get thrown in before the original 6 get paired off?     I'm really confused now.    

oops. I guess I didn't see that. sorry. and thankyou for pointing that out

I'm afraid that my answer might now have been correct, so I deleted it.

If it is an arithmetic sequence, then if you subtract

   the first term from the second (69 - -31)

   and the second term from the third (169 - 69)

   and the third term from the fourth (269 - 169)

you will get the same answer every time ...

Are the answers to these three subtraction problems the same number?

If so, it's an arithemetic sequence; if they are not all the same answer then it isn't an arithmetic sequence ...

(px + q)(2x + 5)  =  6x² + 11x + r

2px² + 5px + 2qx + 5q  =  6x² + 11x + r

--->   2px² = 6x²  --->   p = 3

--->   5px + 2qx = 11x   --->   5p + 2q = 11

         --->  Since p = 3, 5(3) + 2q = 11

         --->   15 + 2q  =  11

         --->   2q  =  -4

         --->   q  =  -2

--->   5q = r

         --->   Since q = -2, 5(-2) = r

         --->   r  =  -10 

$${\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{2}} = -{\mathtt{1}}$$

Assuming that they only vowels are an A on spinner 1 and an E on spinner 2:

There are four possibilities:

spinner 1 = A and spinner 2 = E  =  (3/4) x (1/3) = ?

spinner 1 = A and spinner 2 = not E = (3/4) x (2/3) = ?

spinner 1 = not A and spinner 2 = E = (1/4) x (1/3) = ?

spinner 1 = not A and spinner 2 = not E = (1/4) x (2/3) = ??

If you add the answers to the three possibilities that have a '?', you will get the probaiblity of getting at least one vowel.

You need to use the combination formula:  _nC_r · p^r · q^{n - r}.

n = number of trials = 10

r = number of successes = 0

p = probability of success = 0.90

q = probability of failure = 1 - p = 0.10

I checked what you said and you are right google says it is a graph.  How silly is that.

The graph represents the equation

y=4(x+5) + 3(x+3) 

but that has nothing to do with your questions.   

The width of a dollar bill is 2.61 inches.

One billion dollar bills (side by side) would be 2.61 x 1,000,000,000 inches.

To change these into feet, divide by 12:  2.61 x 1,000,000,000 ÷ 12.

To change that into miles, divide by 5280:  2.61 x 1,000,000,000 ÷ 12 ÷ 5280.

(If you meant end to end, the long way, replace 2.61 inches with 6.14 inches.)