All Answers 356468 Answers

x^2  -  7x   =  38 

x^2  - 7x  - 38  =  0   

Using the Q  formula

x₁  =      (  7  - sqrt  [ 7^2  -  4 (-38)  ]  )  /  2    =    ( 7 - sqrt  [ 201 ] ) / 2   ≈   -3.59

And

x₂ =  ( 7 +  sqrt [201] )   / 2    ≈   10.59

Not  sure  how to solve this algebraically, but  WolframAlpha  gives solutions

x₁   ≈  6.18   =  6.2 ( rounded)

and

x₂  ≈  3.61  =   3.6  (rounded)

Assuming no restrictions  ( some boxes may be empty )  we  have

k=   no. of balls

n = no. of boxes

The  total  ways  =    n^k     =   3⁴  = 81  ways

Monty  must  have painted     2/5   -  1/4   =   (3/20) of the  house

So

(3/20)  /  8   =    3 /160  of the house in one hour

(a)   What is the point estimate of the population?

the point estimate will be equal to the population proportion which is $ 0.62 $ just as stated.

(b)   The margin of error of the population proportion is found using an estimate of the standard deviation. What is the interval estimate of the true population proportion?

just use the standard error formula;

$z \times \sqrt{\frac{p \times (1-p)}{n}}   $

where $p=$point estimate   ;  $n=$sample size ; $z=1.96$

thus:

$1.96 \times \sqrt{\frac{0.62 \times (0.38)}{200}}    $
$ 1.96 \times  0.0343  $

which gives us 
$ 0.0672$ which you can write as $  6.72 \%  $

(c)   The margin of error of the population proportion is found using half the range. 

What is the interval estimate of the true population proportion?

the simulation tells us that the minimum sample percentage is 0.46 and the maximum sample proportion is 0.76

MOE uses half the range(difference) so:

$  \frac{(0.76-0.46) }{2}  $

that gives us exactly $0.15$, and that is the answer.

(d)   Is the community action group’s claim likely based on either interval estimate of the true population proportion? Explain.

yes thats true, as we can see the confidence interval value is greater than 0.45

120 m/s     answered the other day

We know the average flat speed is 120

If you draw a diagram you will see that the up and the down paths are equal for his roundtrip

Call the distance one way = d     2 d for round trip

time on uphill    d/100   time on downhill   d/150

average speed      2 d /(d/100+d/150)  = 2d/ ( 3d/300 + 2d/300) = 2d / ( 5d/300) =  600/5 = 120 m/s

so all of the sections average 120 m/s     so his overall average speed is 120 m/s

I  believe   [ CXYZ ]    will form  an oblique  pyramid

Its  volume is

(1/3) [ XYC ]  (CD  / 2)  =

(1/3) [ XYC ] * 16

(1/6)  [ XYC ]  * 16   =

(1/6)   (1/2) (6)^2 sin (60°) *   16  =

16 * 3 sin (60°)  =

48 sqrt (3)  / 2   =

48 sqrt (3)   units^3

I don't know why.....but.....that's an excellent answer   !!!!

you are absolutely right -- i just perceived like something was off at the moment, dont know why -- wanted to edit the answer, and take the last line off, but it wouldn't let me!

101 ,  197 , 199 , >>>Total sum= 497

i mean $ \overset{-}{AD}=\overset{-}{FE} $

knowing all that we can say that $\overset{-}{DB}$ and $ \overset{-}{DE} $ form an isosceles triangle so yes logically we get $\overset{-}{AB}=\overset{-}{DE}+\overset{-}{FE}=25 $

for the other side we can say that $ \overset{-}{AF}  $ and $\overset{-}{DE}$ have the same measurement -- so does $\overset{-}{FE}$ and $\overset{-}{FC}$ -- thus we can say that $\overset{-}{AC} = \overset{-}{DE} +  \overset{-}{FE}  = 25$ 

so yes, since we have the two sides just add them together 

although may i know why i cannot edit the original answer?

All 4 integers share the following divisors:

(1, 1, 1, 1,    2, 2, 2, 2,    4, 4, 4, 4,    5, 5, 5, 5,    10, 10, 10, 10,    20, 20, 20, 20)

A total of  6 Divisors as listed above.

look at the image -- since $ \overset{-}{AD} = \overset{-}{AC}   $ then we can say that $  \overset{-}{AB}=\overset{-}{DE}+\overset{-}{FE}=25    $

its the exact same from the other point of view, so its P would be $  25+25  $, which is equal to 50

Try typing without $\LaTeX$ (the dollar marks). 

z  = a + bi                        i^2   =  -1

( a + bi)^2    = 2i +  2

a^2  +  2abi   + b^2i^2     =  2i  +  2

a^2 -   b^2  +  2ab i =   2i  + 2

Which implies  that

a^2  - b^2     =   2                (1)  

2ab  =    2    ⇒   ab =  1   ⇒   b =  1/a     (2)

Sub (2)  into (1)

a^2  -  (1/a)^2   =   2       mutiply through by  a^2

a^4   -  1  = 2a^2

a^4  -  2a^2   -  1   =  0               let  a^2  =  x   ⇒   a  = sqrt (x)

x^2  - 2x  - 1    =  0

x^2  - 2x   =  1              complete  the square on  x

x^2  - 2x + 1  =  1 +  1

(x - 1)^2  =  2

x - 1  =  ±sqrt (2)

x = 1 + sqrt (2)                or       1 -  sqrt (2)

Only  the  first  solution  is  valid

a =  sqrt (x)   =   sqrt  (  1  + sqrt (2) )

b =   1  /  sqrt   ( 1  + sqrt (2) )

z =       sqrt ( 1 + sqrt (2) )    +     1/ (sqrt ( 1 + sqrt (2)) i

Approx 986  meters high assuming   l = w = h    and there are no tombs or other hollow spaces inside   LOL

Need more construction details....amount of overlab between layers etc       Square base ?   Triangular base ?

Slope =   ( 1 - 5)   / ( 3 - 5)  =  -4 /  -2   = 2

So

(c - 1)  /  ( 6 - 3)    =   2

(c  -1)  / 3     =   2

c - 1   = 3 * 2

c - 1     =   6

c  = 6 + 1  =  7

If independent   P( D   l  F )  =  P (D)

P (D)  = 48 / 100      

P(D    l  F  )   =  .18 / . 40    =      45  /  100   

Not- independent

Not  = 

The  coordinates of C  are

[  14  + (1/8) (14 - 2)   ,  4  +  (1/8) ( 4  -  - 2)   ]   =

[    14  +     12/8  ,   4    +  6/8   ]  =

[  14  +  3/2     ,   4  +  3/4   ]  =

(  31/2  ,  19/4)

Pop  mean  =  25

Sample mean   =     ( 25 + 10 + 16 + 39 + 27 + 25 + 25 + 32 + 25 + 22  +  28  + 14 )   /  12   =  24

weird glitch cant type the rest of the stuff

I meant to type:

Find the unique triple $(x,y,z$ of positive integers such that x

Sample space =  listing of all possible outcomes

The last  answer  gives us  this

solving it this way is easier, i reckon.

$ \begin{cases} ab^2 = \frac{27}{5} \\  a^2b = 1080  \end{cases} $

lets work this -- $ ab^2=\frac{27}{5} $ ; take the $b^2$ to the other side and you get

$ a=\frac{27}{5b^2}   $

plugging that in the other equation you get \( (\frac{27}{5b^2})^2\times b=1080  \)

that is equal to $   \frac{729}{25b^3} \times b=1080 $

$ 729=27000b^3  $

$  b^3=\frac{27}{1000}$

$ b=\sqrt[3]{\frac{27}{1000}}   $

$  b=\frac{3}{10} $

back to our a equaltion:  $  ab^2 = \frac{27}{5} \ \ \ \Rightarrow \ \ \ \   a (\frac{3}{10})^2 = \frac{27}{5}  $ 

$a\frac{9}{100}=\frac{27}{5}$

$  \cancel{100}a\frac{9}{\cancel{100}}=\frac{27\times100}{5}  $

$  9a=540 $

$  a=60 $

now to our last condition $a+5b   \ \ \ \Rightarrow \ \ \  60+ 5(\frac{3}{10}) $

$  60+  \frac{5\times 3}{10} $

$  60+ \frac{3}{2}  $

$ \frac{60\times 2+3}{2} $

$  \frac{123}{2} $

:D

Thanks CPhill! I appreciate the help