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Oh dear, I seem to have made a typo. For question #1 it is asking for the data between the mean and 0.6. My apologies. Does this mean that your first answer was correct? The one that I also got the same answer for? Sorry again!

Sorry...I'm having a bad day!!!

Look at my revised answer....AGAIN

I made the same mistake originally

z score for  0.6  =  .7257

But  we want the  score associated with  .06

This  is  .5239

Question #2 it is talking about .4 but in your answer, you found the z score for .04?

3) In a standard normal distribution, what percent of the data is between two and three standard deviations below the mean?

The percent of  data  falling between  two and three SDs   below the mean is the same  percent of data falling between two and three SDs  above the mean

The  percent   =   1.7%  between  2  and 2.5 SDs above the mean plus   .5 %  between 2.5 and 3 SDs  above the mean

So....the total %   =   [ 1.7  + .5 ] %  =   2.2%

I have a chart here and for 0.6 it says that 0.6 = .2257. Does that change anything? I am to use the chart that I have.

2) In a standard normal distribution, what percent of the data is between -0.4 and 0.4?

The z score value associated with .0.4  is .6554

The z score  value associated with -.0.4  is .3446

So.....the percent of data falling between these   is [ .6554 - .3446]  = .3108  =  31.08%

CORRECTED ANSWER  !!!!

Sorry....I used the wrong value.....check my revised answer  !!!!

Nice I got the first one correct, although I did it a little differently than you did. 

1)   Here...I'm assuming that  .06   is a z score

The value  associated with a z score of  .06  =  .5239

The  mean will  have a z score of 0    .....of course.....the value associated with the mean  = .5000

So.....the percent of data falling between .5000  and .5239   =

[ .5239 - .5000]  =   ..0239  ≈    2.39%

Here's  one more way to see  this

Let us suppose  that  the result  of taking the square root  is some integer  > 12

Then....when we square both sides  we have that

144 - (x)^(1/3)  =  (minimum of 169)      

Subtract   144 from both sides

-(x)^(1/3)   =   (minimum of 25)

Call the right side some positive integer, P

- (x)^(1/3)  =   P          

(x)^(1/3)  =  -P        

(x)^(1/3) = (-1)P     cube both sides

x = (-1)^3 P^3

x =  - P^3

So.....x  is negative.....but  we require that x  is  positive....so.....the result of evaluating the original expression cannot be an integer > 12

Now suppose that   is  some  integer  between 0 - 12   inclusive

So.....when we square  both sides......the max  = 144   and the min  = 0

If the max is 144 then

144 - (x)^(1/3)  =  144

-(x)^(1/3)   =   0

x^(1/3)  =  0

And x  =  0       so  x is non-negative

If the min is  0

144 - (x)^(1/3)   =  0

-(x)^(1/3)  = - 144

x^(1/3)  =   144

x  = 144^3    and x is also non-negative

So x will  be positive   whenever  the original quantity under the root is a perfect square and the right side  is an integer from  0 to12   inclusive

Well....since you asked....

   3x  −y = 6

   6x + y = 21                add these equations

9x    =    27               divide both sides by 9

x  =   3

Using either equation to find y

3(3) - y  = 6

9 -  y  =    6         add y to bothsides, subtract 6 from both sides

3   =   y

So  (x, y)   = (3, 3)

\(\tan \theta \sec \theta = 1,find \frac{1 + \sin \theta}{1 - \sin \theta} - \frac{1 - \sin \theta}{1 + \sin \theta}. \)

[ I'm using  A  instead of  Theta ]

1 +  sin A           1  - sin A

_______    -    __________        get a common denominator

1  - sin A            1 +   sin A

(1 + sinA)^2  - (1 - sin A)^2

_______________________

(1 + sin A) (1 - sin A)

(sIn^2A  + 2sin A  + 1)   - ( sin^2A - 2sin A + 1)

_______________________________________

               1  -  sin^2 A

2 sin A  +  2 sin A

_______________

        cos^2 A

  4 sin A

_______

 cos^2 A

    4 sin A

____________

cos A * cos A

4 * sin A  *   1

    _____    ____

     cos A    cos A

4   ( tan A   sec A )

4     (1)   =

4

Second one

1      2       -7                  Multiply R1 by 3  add to R2

-3   -5       23

1     2      -7                 Multiply  R2  by  -2   add to R1

0     1       2

1      0     -11

0      1       2

Read the last column down and we have that  (x, y)  =  (-11, 2)

First one

1   2   1    5

2  -1   2   15             Multiply R1 by -2    add to R2

3   1  -1    8

1   2    1    5

0   -5   0    5           Multiply R1 by  -3    add to R3

3   1   -1    8

1  2    1     5

0  -5   0     5               Multiply R2  by  -1/5

0   -5  -4   -7

1   2   1     5

0   1    0    - 1     Multiply R2  by  -2    add to R1

0   -5  -4   - 7 

1    0     1     7

0    1     0   -1       Multiply  R2 by 5    add to R3

0   -5    -4   -7

1     0    1     7

0     1    0    -1        Multiply  R3  by  -1/4

0     0   -4   -12

1   0    1     7

0   1    0    -1       Multiply  R3   by -1  add to R1

0   0    1     3

1     0    0     4

0     1    0    -1

0     0    1     3

Reading the last column  down  ( x, y, z)   =  (4, -1, 3)

Thank you Melody, that is a really cool keychain!!

Melody I am sorry for being so rude. Thank you rom, and Thank you melody for answering. question 3 is not the same as 1, the game ends after 1 side reaches 4 points.

No graph shown...

An example of a graph would be (x+3)(x-2)(x-5).

Find the number of ways of arranging the numbers 1, 2, 3, 9 in a circle, so that the sum of any three adjacent numbers is divisible by 3. (Two arrangements are considered the same if one arrangement can be rotated to obtain the other.)

There are none.

Rom I think you misunderstood the question

Patricia's investment =$50,000 x 1.01^8 =$54,142.84

Jose's investment       =$50,000 x 1.04^2 =$54,080.00

The difference           =$54,142.84 - $54,080.00 =$62.84 - Extra money that Patricia's investment earned.

sweet