All Answers 356333 Answers

First group them into a frequency-distribution table

                x                                   f                                   c.f                            c.f (without outliers)

               20                                  1                                   1                                       

               60                                  1                                   2                                             1

               65                                  2                                   4                                             3

               70                                  2                                   6                                             5

               75                                  4                                  10                                            9

               80                                  1                                   11                                           10

               85                                  2                                   13                                           12

               90                                  1                                   14                                           13

               95                                  1                                   15                                           14

              229                                 1                                   16

                                                ∑f = 16

a) Mean \(={20+60+65(2)+70(2)+75(4)+80+85(2)+90+95+229\over 16}\)

    Mode = 75 

    Median is at (16/2)^th and (16/2+1)^th positions i.e 8^th and 9^th positions. 

    Median \(={75+75\over 2}\) \(=75\)

b) The outliers are 20 and 229. 

c) Without outliers, 

    Mean \(={60+65(2)+70(2)+75(4)+80+85(2)+90+95\over 14}\)

    Mode = 75

    Median is at 7th and and 8th positions

    Median \(={75+75\over 2} = 75\)

   How is mean affected? Calculate the difference between both the means. 

d) Now you might as well think about that one... its easy, would you rather pay less or pay more? 😏 

Just as Jugoslav pointed out, there's a mistake in this question; namely, instead of 18 cm, should be 180 cm!

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The total length of the lines in the 1st square is 180 cm.

The total length of the lines in the second square is 72 cm.

I've already given you some obvious hints.  You should be able to figure out from there.

Solve \(z^3 = 8*i\) in complex numbers.

\(\begin{array}{|rcll|} \hline z^3 &=& 8*i \quad | \quad \text{cube root both sides} \\ z &=& \sqrt[3]{8i} \\ z &=& \sqrt[3]{8}\sqrt[3]{i} \\ \mathbf{z} &=& \mathbf{2\sqrt[3]{i}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline i &=& e^{i\frac{\pi}{2}} \\ \sqrt[3]{i}=i^{\frac{1}{3}} &=& \left(e^{i\left(\frac{\pi}{2}+ 2\pi*k \right)}\right)^{\frac13} \\ \mathbf{\sqrt[3]{i}} &=& \mathbf{e^{i\left(\frac{\pi}{6}+\frac{2\pi*k}{3}\right) } } \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline k=0: & \sqrt[3]{i} &=& e^{i \frac{\pi}{6} } = \cos(30^\circ)+i*\sin(30^\circ) \\ k=1: & \sqrt[3]{i} &=& e^{i \left( \frac{\pi}{6}+\frac{2\pi}{3}\right) } = \cos(150^\circ)+i*\sin(150^\circ) \\ k=2: & \sqrt[3]{i} &=& e^{i \left( \frac{\pi}{6}+\frac{4\pi}{3}\right) } = \cos(270^\circ)+i*\sin(270^\circ) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{z} &=& \mathbf{2\sqrt[3]{i}} \\ \hline z_1 &=& 2*\left( \cos(30^\circ)+i*\sin(30^\circ) \right) \\ z_1 &=& \sqrt{3} + i \\\\ z_2 &=& 2*\left( \cos(150^\circ)+i*\sin(150^\circ) \right) \\ z_2 &=& -\sqrt{3} + i \\\\ z_3 &=& 2*\left( \cos(270^\circ)+i*\sin(270^\circ) \right) \\ z_3 &=& -2i \\ \hline \end{array}\)

Why don't YOU post what YOU have tried? 

You can take a photo of your diagram and upload it.

one question per post and one question at a time.

It looks like you are just asking people to do your homework for you.

Aww, thank you. :))

=^._.^=

_ _ _

last digit 2 options (2, 4)

first digit 3 options (4, 3, 2, 1, but not 2 or 4), can't be 5 since that would be larger than 500. Minus 1 from the last digit. 

middle digit 3 options. can't be first/last digit. 

2*3*3 = 18

=^._.^=

Ok I have commented on the other thread.

I accept your apology but I do not think that you actually read my response at all.

You certainly have not attempted to answer any of my questions.

ok, perhaps you have a major problem with reading and interpreting questions. Although I doubt you seriously read my response.

I did not ask you how to get to the final answer.

I asked you to tell me the initial conditions.

I asked you for the initial vertical speed, the initial horizontal speed, the initial acceleration, the initial vertical and horizontal displacement.

You still have not given me those. Nor have you discussed them in any way.

You cannot seriously understand your original question if you cannot answer my questions. 

Se seu amigo tirar um terço de suas maçãs, então você tem dois terços de suas maçãs.Dois terços de 999 é 6666666.

(a+bi)^3 = 8i

-b^3*i - 3ab^2 + 3ab^2 * i + a^3

a^3 - 3ab^2 = 0

3ab^2 - b^3 = 8

Now I'm stuck. 

Good luck, I hope you can figure it out. 

Sorry I wasn't able to help more. 

=^._.^=

7 out of (3+7)  is yellow     7/10

     Her new mixture will be  5  out of 7    or  5/7 yellow

7/10 ( 4 gal)     +   y gallons =    5/7 ( 4+y)

28/10

28/10    + y = 20/7 + 5/7 y

196/70   + y = 200/70 + 50/70 y

-4/70 = -20/70 y

y = 4/20 = 1/5 gallon

Take square root of both sides

x-1 = +-2

so x -1 = 2     or  x-1 = -2

     x = 3         or    -1

Sorry....as I noted in my answer above....147 coins originally is incorrect because   147 + 53 = 200      and you have to have MORE THAN 200 coins after adding the 53 .......   

A right triangle has legs of length  6 and b, and a hypotenuse of length c.                                                                      The perimeter of the triangle is 15. Compute c.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

a = 6             b + c = 9           b = 9 - c           P = 15

a² + b² = c²

6² + (9 - c)² = c²

c = 6.5                (b = 2.5)

Sorry, that's incorrect.

The number of ways is c(12,5) - c(7,2)*c(5,2) = 582.

      2

________   =

∛4  + ∛32

       2

___________  = 

∛4  +  2∛4

       2                          ∛ 2

_____________   *   ____    =

      3∛ 4                     ∛2

2 ∛2

______    =

 3 ∛8

2 ∛ 2

_____   =

3 * 2

     ∛2

_______

      3

A + B    =  5

Let  x  be  the  number of  coins in each of  the  7  bags originally

So

7x  + 53  > 200

7x > 147

x >  21

Since we have  8 bags, then  the  number of  coins must  be  an even number  >  200

Through  a little trial  and error, I  find  that  the number of  coins =  256.....and there were 29 coins  in each  of the 7 bags originally

Logic :

Let  29  be  the  number of  coins in each of  the  7 bags originally

Distribute  3   coins  out of  the  53   into each of the  7  bags

Then  each of the  7  bags contains 32 coins   and  the  8th bag contains  53 - 7(3)  = 53 - 21  =  32 coins

So....as found previously.....we must have had    29 x 7  =   203  coins originally

Is there a solution to problem 1 without inversion? I don't really know inversion, like is there an angle chasing solution? If so can you post a solution?

Alex needs 10 quarts of 50%      ....so he needs  5 quarts of antifreeze  and 5 qts water

he currently has only   4 quarts antifreeze   and 6 quarts of water

....a 40% solution

.40 (x)   +   1 (10-x)   =  .50 (10)

.40x + 10 -x = 5

-.6x = -5

x = 5/.6 = 8.333 quarts  of   40%    will be needed to start    then 1.67 quarts of pure antifreeze will be added

10 - 8.33 = 1.67 quarts to drain and refill with 100% a-freeze

80/l   * x      +      125/l * ( 20-x)  = 109 (20)

80x  + 2500- 125x = 2180

- 45x = -320

x = apple juice liters = 7.11        cranberry = 12.89 liters