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I don't  know  two interchangeable terms, but these are examples of  "proper fractions"   where the  numerator  [ top number]  is  less than the denominator  [ bottom number ]

The top number is the NUMERATOR

the bottom number is the DENOMINATOR

I think this is correct.....but someone else might  know better than me !!!

On any  roll of one die...the chance of any number appearing is  (1/6)

So...the probability  of  any  particular arrangement showing up  by  rolling 10 dice is just   (1/6)^10  ≈  0.0000000165381717

Last one

P(A)  = 60 / 150 =   6/15  =  2/5

P(B) =  50 / 150  =  5 / 15  =  1/3

P(A and B) =  (2/5)(1/3) =  2 / 15

P(A l B)  =   P (A and B)                (2/15)

                  __________   =         _______   =       (2/15) * (3/1) =   6/15 =   2/5

                      P(B)                          (1/3)       

P(A l B )  = P(A)  is true.....the events are independent

I can do a couple of them.....

Flip a coin....roll a 6 sided die

P ( Heads  and < 6)  =   (1/2) * (5/6) =    5 / 12

An object has a mass of 0.124kg and a volume of 1893mm3. What is its density in grams per cubic centimeter?

Hello Guest!

\({\color{blue}\rho} =\frac{m}{V}=\frac{0.124kg}{1893mm^3}\times \frac{1000g}{kg}\times\frac{10^3mm^3}{cm^3}\color{blue}=65.5044902272\frac{g}{cm^3}\)

The density of the object is approximately equal to 65.504 g / cm³

  !

No problem.....multiple answers are always good to see other perspectives and ways to solve....and to keep us all honest!

Sorry EP. Didn't see you working on it.

It will be 8:15

1.75 % annual interest is   .0175/12 = .001458333 % monthly (periodic) interest

20 years = 240 periods (months)

FV = 80 000   (future value)

PV = amount to deposit

FV = PV (1+i)^n         n= periods (240)

80000 = PV (1.001458333)^240

PV = $ 56,389.42

One is \(x\times y =60\)

A1) We want \(g^2-2g-24≠ 0\) so we will find when \(g^2-2g-24= 0\) so  \(g = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(g1=\frac{2+\sqrt{2^2-4(1)(-24)}}{2(1)}\)<=>\(g1=\frac{2+\sqrt{100}}{2}\)=\(\frac{2+10}{2}=\frac{12}{2}=6\) and

\(g2=\frac{2-\sqrt{2^2-4(1)(-24)}}{2(1)}\)<=>\(g2=\frac{2-\sqrt{100}}{2}=\frac{2-10}{2}=\frac{-8}{2}=-4\)

Finally \(g1=6,g2=-4\) so We want \(g^2-2g-24≠ 0\) so \(g≠ 6,g≠ -4\)

A2) \(\frac{(x+9)(x-2)}{(x-2)(2+x)}=\frac{(x+9)}{(2+x)}\times\frac{(x-2)}{(x-2)}=\frac{(x+9)}{(2+x)}\times1=\frac{(x+9)}{(2+x)}\)Correct the 3rd answer BUT we assume that the equation is well defined and \(x≠ 2\)

A3)\(\frac{9-x^2}{9x+27}=\frac{(3+x)(3-x)}{(9)(x+3)}=\frac{(x+3)(3-x)}{(9)(x+3)}=\frac{3-x}{9}\times\frac{(x+3)}{(x+3)}=\frac{3-x}{9}\times1=\frac{3-x}{9}\) BUT  like before we assume that the equation is well defined and \(x≠3\)

Hope this helps!

Here is a simple scatter plot I made in Google Sheets. 

A)

B)

I do not see any areas that one can call a cluster. If you cannot identify any clusters or correlations in the graph, then there cannot be any outliers. 

The greatest possible value of GCD(n +7), (2n + 1) is when:

(n +7) =(2n +1), solve for n

2n - n =7 - 1

n= 6. So, GCD (6 + 7), (2*6 + 1) =GCD(13 , 13) =13 - Greatest possible value of GCD.

a. The top of the graph illustrated has a value of y = 4, so you need to solve the equation (x-2)² - 1 = 4 (this is a quadratic equation and will produce two values for x). 

b. Assuming this means the intersection of the curve with the horizontal line at y = 5, there will be two points of intersection.

c.  Since 'a' gives the y-coordinate of the vertex (lowest point), then when a>0 the curve will not intyersect the x-axis.

If 1,000 mod n =1, where n=1, 2, 3,4...........998, 999, 1000, just how many n are there between 1 and 1,000?

\(\text{Since $1000 \pmod 1 = 0$ and $1000 \pmod {1000} = 0$, $n$ can only be $2,3, \ldots , 999$.} \)

\(\begin{array}{|rcll|} \hline 1000 & \equiv & 1 \pmod n \\ \text{or} \\ 1000-1 &=& n\cdot m,~ \text{with } m \in \mathbb{Z} \\ 999 &=& n\cdot m,~ \text{so $n$ are all divisors of $999$ except $1$ } \\ \hline \end{array} \)

The divisors of 999 are:

Divisors:

1 | 3 | 9 | 27 | 37 | 111 | 333 | 999 (8 divisors)

n = 3 | 9 | 27 | 37 | 111 | 333 | 999 (7 numbers)

If it is of the form   f(x) = x k  

then   18=2k     and  k= 9

conversely, it COULD be  yk = x

            then  18k = 2     k = 1/9                 but convention would likely make it the FIRST answer   k= 9

The vertex  is  ( 4, 2)

We can start with this form

4p ( y - k)  = ( x - h)^2     where p  = 1  and ( h, k)  is the vertex

4p ( y - 2)  =  ( x - 4)^2   

4 ( y - 2)   = x^2 - 8x + 16

4y - 8  =  x^2 - 8x + 16

4y  =  x^2 - 8x + 24          divide both sides by 4

y = (1/4)x^2  - 2x + 6 

a = (1/4)      b  =  -2   and  c   =  6

And   their sum is     4 + 1/4   =    17  / 4  

Ahaaaaa!   It is all clear to me now !   Haha       

.124 kg = 124 gms

1893 mm^3 = 1.893 cm^3

124gm/1.893cm^3  = 65.5 gm/cm^3  

Thanks, EP....here's a slightly easier way

2^x  = 7

2^(2x) =

 [ 2 ^x ] ^2  =

7^2   =

49

I think this is supposed to be

-5x - 3y  =  12

y = x - 12

Sub the second equation into the first for y and we have

-5x - 3 ( x - 12)  = 12  simplify

-5x  - 3x + 36  = 12     

-8x + 36  = 12    subtract  36 from both sides

-8x  =  - 24    divide both sides by -8

x  = 3

And    y  =    3 -  12  =  -9

So you are correct!!!!....    (3 , -9)  is the solution

Length of diagonals       sqrt(12^2 +9^2)     and  sqrt(6^2 +8^2)

                                       sqrt  225               and   sqrt 100

                                                 15                             10

Area is 1/2 product of diagonals (given)       1/2 (15 * 10) = 75 sq units

HelpPls.... the file is not opening for me.....

Nor can I access the image adress

Triangle on the left hypotenuse  =  sqrt [  5^2  + (15 - 3)^2  ] =  sqrt [ 5^2 + 12^2 ] =

sqrt [ 25 + 144 ]  = sqrt [ 169 ]  = 13 ft

Triangle on the right  hypenuse =  sqrt [ (14 - 5)^2  + (15 -3)^2 ]    sqrt [ 9^2 + 12^2 ]  =

sqrt [ 81 + 144]   =  sqrt [ 225]  = 15 ft

So...the rope length is   13 + 15  =   28 ft