Three machines operating independently, simultaneously, and at the same constant rate can fill a certain production order in 36 hours. If one additional machine were used under the same operating conditions, in how many fewer hours of simultaneous operation could the production order be fulfilled?

A. 6
B. 9
C. 12
D. 27
E. 4

D

Step-by-step explanation:

Let the total production order be X. The combined rate is thus x/36 orders per hour.

Now, we know that the three machines are working at the same constant rate. This means that individual rate for each of the machines will be x/36 divided by 3 and that gives x/108 per machine.

Now, we are having another machine coming at the same constant rate. This means we are adding an x/108 rate to the preexisting x/36.

The new total rate thus becomes x/108 + x/36 = 4x/108

Now we know that the total new rate is 4x/108. Since the total work doesn’t change and it is still x, the time taken to complete a work of x orders at a rate of 4x/108 order per hour would be x divided by 4x/108 and this is x * 108/4x = 108/4 = 27 hours

b

step-by-step explanation:

the total is 23.97. 0.25% is off so you have to subtracting 25% from 100% and you will get 75% if you convert it to decimal you will get 0.75. you have to find the original because the formula is % times original = total.so you have the total and the % so you just have to plug it in.

use an equation of ratios to determine the amount of gas needed to fill the tank.

    0.2             0.8

  =    

288.75             x

                                                              231 gal

so, 0.2x = (0.8)(288.75 gal), and x = = 1155 gal.

                                                                0.2

it would take 1155 gal. to fill the tank.

the cost would be

1155 gal         $3

* = $3465   (answer)

      1           1 gal

hope this you