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6. In a bacteria colony consisting of 210,000 bacteria, [tex]15\%[/tex] of the bacteria have Mutation A, [tex]\frac{2}{3}[/tex] of the bacteria have Mutation B, and the rest of the bacteria have no mutations. How many bacteria in this colony have no mutations?

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GinnyAnsweridn
Certainly! Let's solve this step by step.

### Step 1: Total Number of Bacteria
We know that the total number of bacteria in the colony is:
[tex]\[ \text{Total bacteria} = 210,000 \][/tex]

### Step 2: Calculation of Bacteria with Mutation A
We are told that 15% of the bacteria have Mutation A.
To find out how many bacteria have Mutation A, we calculate:
[tex]\[ \text{Bacteria with Mutation A} = 0.15 \times 210,000 \][/tex]
[tex]\[ = 31,500 \][/tex]

### Step 3: Calculation of Bacteria with Mutation B
We are also informed that [tex]\(\frac{2}{3}\)[/tex] of the bacteria have Mutation B.
To determine the number of bacteria with Mutation B, we compute:
[tex]\[ \text{Bacteria with Mutation B} = \frac{2}{3} \times 210,000 \][/tex]
[tex]\[ = 140,000 \][/tex]

### Step 4: Calculation of Bacteria with No Mutations
The rest of the bacteria have no mutations. To find out how many bacteria have no mutations, we subtract the number of bacteria that have Mutation A and Mutation B from the total number of bacteria:
[tex]\[ \text{Bacteria with no mutations} = 210,000 - \text{Bacteria with Mutation A} - \text{Bacteria with Mutation B} \][/tex]
[tex]\[ = 210,000 - 31,500 - 140,000 \][/tex]
[tex]\[ = 38,500 \][/tex]

### Conclusion
The number of bacteria in the colony that have no mutations is:
[tex]\[ 38,500 \][/tex]
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