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Let g(x) = y

These are easy to solve....the product of x and y in the inverse variation is constant

So...

y * x   = (0.2) (0.1)  = .02

So....when x = 1.6

y * 1.6  = .02       divide both sides by 1.6

y = 0.0125 = g(x)

y+4= 2/5(x−3)    simplify

y + 4 = (2/5)x - 6/5     subtract 4 from both sides

y = (2/5)x  - 6/5 - 4

y = (2/5)x - 6/5 - 20/5

y = (2/5)x -  26/5

The y intercept is  (0, - 26/5)    plot this point

The slope is 2/5.......go up 2 on y and to the right 5 on x.....this gives the point 

(0 + 5, -26/5 + 2)  =  (5, -26/5+ 10/5)  =  ( 5, -16/ 5)   plot this point

Draw a  line through these points.......that will be your graph

y + 4 = 25 ( x - 3)   simplify

y + 4  =  25x- 75      subtract   4 from each side

y = 25x - 79

When x = 0   y =  25(0) - 79  =  - 79

When x - 1, y = 25(1) - 79 =  - 54

So.....the points   (0, -79)  and (1, -54) are on the graph

Plot these two points......draw a line through them.....that will be your graph....!!!

The inverse reverses the  original coordinates

Since   (1, -5) is n the original graph....then  (-5, 1) is on the inverse ⇒  "D"

Lot of questions on here, JGA.....hard to find any particular one....

Thanks CPhill, don't know how I didn't see that..

 https://web2.0calc.com/questions/help_53845

The mean is 82.2

The standard deviation is  ≈ 9.8

So...the srcores that fall within one SD  is

82.2 - 9.8  ≈ 72.4   and

62.2 + 9.8 ≈ 92

Note that 7 scores fall within this range  = 7/10  = 70%

This is close to expected  [ we expect 68%]

"C"

The slope=$\frac{y_2-y_1}{x_2-x_1}=\frac{y-5}{8-(-2)}=\frac{y-5}{10}=\frac{-1}{5}$, so $5y-25=-10, 5y=15, \boxed{y=3}.$

B.

The majority of the data is to the right of the mean.

68%  of the values  fall within 1 standard deviation of the mean

So....the range is  given by

60 - 4.5  =  55.5 in    

and

60 + 4.5 = 64.5 in

"B"

log ₃ [3^ (-4)]   = 

log ₃ 3^(-4)  =

(-4) log ₃ 3            {log _a a   =  1   }

-4 * 1   =

-4

Question  1

Just double the x and y coordinates for each corner of the picture and the mat and plot these new points

For instance....the top rigt corner of the mat becomes (2,3) *2  = (4, 6)

And the top right corner of the picture = (1,2) * 2  =  (2, 4)

etc.

Question 3

a)  The perimeter of each new image = 2 times the perimeter of the old image

     The area of each new image =   (scale factor)^2 * area of old image = 4 times the old image

b)  Scale factor of 3 triples the perimeter  and the area is (3)^2 = 9 times as much

     Scale factor of 4 quadruples the perimeter and the area is (4)^2 = 16 times as much

P (syrup l  waffles ) =  P (syrup and waffles) / P (waffles)  = 0.3 / 0.5 =  3/5 = 0.6

This =  P(syrup)

So

Independent

P ( male and Brand A )  =  P( male l Brand A ) * P (Brand A) =  (.36/.75) * (.75)  = .36  = P(male)

Independent     →  First box

=      ⇒   Second box

1.The probability of drawing two red cards without replacement is 25/102 , and the probability of drawing one red card is 1/2 .

What is the probability of drawing a second red card, given that the first card is red?

P ( 2nd red l 1st red)  =   P ( both red) / P( 1st red)  =  

(25/102) / ( 1/2)  =  50/102 = 25/51

2.The probability of choosing two green balls without replacement is 1/11 , and the probability of choosing one green ball is 1/3.

What is the probability of drawing a second green ball, given that the first ball is green?

P ( 2nd green l 1st green )  =  P( both green) /P(1st green)  =  (1/11) / (1/3) = 3/11

3.A math teacher gave her students two tests. On the first test, 75% of the class passed the test, but only 60% of the class passed both tests. What is the probability that a student passes the second test, given that they passed the first one?

P ( passing second l  passed 1st)  =  P(passing both) / P (passing 1st) =

.60 / .75  =  4/5  =  .80

For the value of 'y' to equal ZERO (all of the points are x-intercepts and thus = 0 )  i.e.  f(x) =0

then   "D"

f(x)= 0 =  (x+1) (x+2)(x-3)     So at least ONE of the factors must equal 0

               (x+1) = 0      would mena x = -1

                         (x+2) = 0    would mean   x = -2

                                 (x-3) = 0 would mean    x = 3      so answer D

I have finally solved the problem. Oh, you changed the problem! So, is the answer $\boxed{-\frac{137}{60}\quad}.$

Now, the LATEX: $\left(\frac{\left(-\frac{3}{5}+\frac{1}{10}\cdot \:3\right)}{\left(\frac{\frac{9}{20}}{\left(-\frac{1}{4}-\frac{1}{5}\right)}\right)}\right)-\frac{3}{4}-\frac{\frac{2}{5}}{\frac{\left(\frac{1}{20}-\frac{1}{4}\right)}{\left(-\frac{5}{6}-\frac{1}{13}\left(\frac{5}{4}-\frac{1}{6}\right)\right)}}$

The point is given by

[ (9,13) - (4,1) ] (4/5) + (4,1)  =

[ 5, 12] (4/5) + (4,1)   =

(4, 48/5) + (4,1)  =

(8, 53/5)

Thanks cphill

Okay

All problems can be handy by learning the Principle of Inclusion-Exclusion formula. Look at CPhill's solution for more detail!

1.Cathy rolls a number cube that is labeled from 1 to 6. What is the probability that she rolls an even number or a multiple of 3?

P (even or multiple of 3)  = P(even) + P(multiple of 3) - P(even and multiple of 3)  =

                                             3/6      +       2/6       -     1/6  =

                                                            4/6   =     2/3

2.Joseph has 20 coins in his pocket. He has 5 nickels, 6 quarters, 7 dimes, and 2 pennies. Joseph chooses a coin at random.

What is the probability that he selects a quarter or a penny from his pocket?

P ( Q or P )  = P(Q)  + P(P)  -  P (Q and P)  =

                        6/20    +  2/20  -   0/20     =   8/20    =   2/5

3.There are 20 cards labeled from 1 to 20 in a bag. A card is randomly selected from the bag.

What is the probability of getting an even number or a 1?

P (E or 1)  =  P(E)  + P(1)  - P( E and 1)  =

                      10/20 + 1/20  - 0/20    =     11/20

2. Remember that OR=addition, so we have 6/20+2/20=8/20 or 2/5.