Answer:
so cant help
Step-by-step explanation:
Answer: Carol bought 4 pounds(lb) of meat.
Explanation: 0.32 x 5 = 1.60, and 0.8 x 3 = 2.40, and 1.60 + 2.40 = 4.
There would probably be multiple ways to solve this, but I will only show you one of them.
First, we need to figure out how many tablespoons of the chemical solution it takes to decrease the number of bacteria in a gallon to 300. So we subtract 1200 by 300, to find the number of unnecessary bacteria, which is 900. Now we divide 900 by 75, which is 12. So it would take 12 tablespoons of the chemical solution to make a gallon have 300 bacteria.
There are 500 gallons, and we need to put 12 tablespoons of the chem-solution in each gallon. So we multiply 500 by 12, which is 6,000. Meaning that 6,000 tablespoons of chemical solution would be needed to make the 500-gallon tank have 300 bacteria per gallon.
Answer:
-19.
Step-by-step explanation:
-19+31 gets you 12.
The missing values in the scales given are :
- 20; 40; 80
- 15; 30; 45, 75
- 25; 50; 75; 125
- 50; 150; 200; 250
Each successive unit on a graph represent a specific value :
The time in seconds represented by each unit:
- Between 0 to 60 seconds ;
- Number of units = 3
- Number of seconds per unit = 60 / 3 = 20 seconds
- Each unit represents 20 seconds
- The missing units are : 20; 40; 80
For the graph in (b)
The time in seconds represented by each unit:
- Between 0 to 60 seconds ;
- Number of units = 4
- Number of seconds per unit = 60/4= 15 seconds
- Each unit represents 15 seconds
- The missing units are : 15; 30; 45, 75
For the graph in (c)
The time in seconds represented by each unit:
- Between 0 to 100 seconds ;
- Number of units = 4
- Number of seconds per unit = 100/4= 25 seconds
- Each unit represents 25 seconds
- The missing units are : 25; 50; 75; 125
For the graph in (b)
- The time in seconds represented by each unit:
- Between 0 to 100 seconds ;
- Number of units = 2
- Number of seconds per unit = 100/2 = 50 seconds
- Each unit represents 50 seconds
- The missing units are : 50; 150; 200; 250
Hence, the procedure for calculating units values of a graph.
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