CPhill

Second one

This graph looks to have a moderate  positive correlation

First one

Plotting the training hours on x  and tthe accidents on y we have the following points

(0,8)  (12,0)  ( 2,5) (10, 1) (9, 3)  (5,2) (6,2) ( 4, 0)

See the graph below

The r value   =  -.7283   .....this indicates a strong negative correlation.....in other words.....the fewer the training hours, the more accidents

x^2 - 3x + 7  = 5 - x^2         add x^2 to both sides......subtract 5 from both sides

2x^2 - 3x + 2  =  0        (1)

The dicriminant   is

(-3)^2 - 4(2)(2)  =

9  -  16  =

-7

Thus.....if the discriminant is negative, we have no real solutions to   (1)....so.....there are no intersection points  as shown by the graph here :

https://www.desmos.com/calculator/1gyqxrcryq

Measure of angle QTR  =  (1/2)arc QR  = (1/2)(110)  = 55°

Measure of angle RQS  = (1/2) arc RS  = (1/2)(94)  = 47°

Measure of arc TS  =  [ note that arc QT  =  2 m angle ORT  =  2*27  =  54°

So  measure of arc TS  =

360 - arc QR - arc RS - arc  QT  =

360 - 110 - 94 - 54  = 

102°

Measure of angle TRS  =  (1/2)arc TS  =  (1/2)(02)  = 51°

Measure of angle QSR  =  measure of angle QTR  = 55°

Let's see if we can get a larger answer.....

Mean  = 6

Mode = 5

Median = 4

 1  2   3  4  5  5  22

Yes!!!.....you are correct,  Nirvana.....we CAN do better than 15  !!!!

Good, job!!!!

Matrix P  =                                               Matrix  I  =    

[  3  4    1    1 ]                                               [ 3  ]

   6  6    4    6                                                   5

[  2  2    5    4 ]                                                 2

                                                                      [  6 ]

Matrix  PI  =

[  3*3 + 4*5 + 1*2 + 1*6 ]              [  9 + 20 + 2 + 6   ]              [ 37 ]

   6*3 + 6*5 + 4*2 +  6*6        =       18 + 30 + 8 + 36        =       92 

[  2*3 + 2*5 + 5*2  + 4*6 ]             [  6 + 10 + 10 + 24 ]             [ 50 ]     

OK!!!!

Good job, CU   !!!

Note that  ( .125)   =    125/1000  =   [25 * 5] / [ 200 * 5 ]  =  25 / 200   =   1 / 8 

So  we have  (1/8)^3  = ( 1/2^3)^3  = 1/2^9

1 / 2^9      =    [ remember that   1/ b^n  =  b^(-n)  ]....so..... 

1/2^9  =   2^(-9)

So  "a"  = -9

Here's my attempt  !!!

Mean  =   sum / number of integers  =  42/7  =    6

We would like to make the mode  (and median) as small as possible.....

Let the mode = 4  and the median   = 5

And we want the first  integer to be as small as possible  = 1

1  4  4  5  6  7 15

The mode  = 4

The median  = 5

The average = 6

I get 15  as the largest value.......???

Maybe CU  has a better answer    !!!