The number of bacteria in a petri dish on the first day was 113 cells. If the number of bacteria increase at a rate of 82% per day, how many bacteria cells will there be after 7 days?

4107 cells

Step-by-step explanation:

From the question, we have the following values:

Day 1 : 113 cells

Number of cells increases by day by 82%

Hence,

Day 2

113 × 82% = 92.66cells

Hence, Total number of bacteria cells for Day 2 = 113 + 92.66 = 205.66cells

Day 3

205.66 × 82% = 168.6412 cells

Hence, Total number of bacteria cells for Day 3 = 168.6412 + 205.66 = 374.3012 cells

Day 4

374.3012 × 82% = 306.926984 cells

Hence, Total number of bacteria cells for Day 4 = 306.926984 + 374.3012 = 681.228184 cells

Day 5

681.228184 × 82% = 558.60711088 cells

Hence, Total number of bacteria cells for Day 5 = 558.60711088 + 681.228184 = 1239.8352949 cells

Day 6

1239.8352949 × 82% = 1016.6649418 cells

Hence, Total number of bacteria cells for Day 5 = 1016.6649418 + 1239.8352949 = 2256.5002367 cells

Day 7

2256.5002367 × 82% = 1850.3301941 cells

Hence, Total number of bacteria cells for Day 7 = 1850.3301941 + 2256.5002367 = 4106.8304308 cells

Approximately to nearest whole number, the total number of bacteria cells that would be present after 7 days = 4107 cells

multiplication is multiplication.   it might feel harder depending on what level of math you are in, but generally it will be the same old stuff, while implementing some new concepts and methods of solving work.