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allochka39001 [22]
3 years ago
11

A bacteria population triples in number each day . if there are 2,551,500 bacteria on the 7th day , how many bacteria were prese

nt on the first day ?
Biology
2 answers:
Gre4nikov [31]3 years ago
8 0
<span>the total number of bacteria present on first day would be

2,551,500=x*<span>3^6
</span></span><span>x=<span><span>2,551,500 / </span><span>3^6
</span></span>=3,500</span>
Sedaia [141]3 years ago
4 0
Well it is triples you can tell that it is a division question.  So triples means by 3. 
The equation is 2,551,500 = d X 3^6. So, now 2,511,500 / 729 is your final answer, 3,500. Hope it helps :)


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What happens to the carbon dioxide (CO2) levels of a plant when it does or does not have light?
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2 years ago
In your own words, explain the difference between the pitch and intensity of sound.
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6 0
1 year ago
Calculate the final concentration of BSA in the problems below using the formula
Rudiy27

Answer:

A. C_2=1.5\frac{mg}{mL}

B. C_2=0.075\frac{mg}{mL}

C. C_2=0.01\frac{mg}{mL}

D. C_2=0.001\frac{mg}{mL}

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

C_2=\frac{C_1V_1}{V_2}

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}

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7 0
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