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Svetllana [295]
3 years ago
7

scientists believe the earth likely warmed between 0.65 celsius and 1.05 celsius with an average value of 0.85 celsius between y

ears 1880 and 2012. what was error bar for the increase in temperature over that period?
Biology
1 answer:
scoray [572]3 years ago
4 0

Answer:

The lower limit is 0.65°C

The upper limit is 1.05°C

the average is 0.85°C

The difference between the average and the lower limit is:

0.85°C - 0.65°C = 0.20°C

the difference between th average and the upper limit is:

0.85°C - 1.05°C = -0.20°C

Then we can write the change in temperature as:

T = 0.85°C ± 0.20°C

Now we want to find the error bar.

The error bar would be the total distance between the lowest value and the largest value.

Then the error bar is:

1.05°C - 0.65°C = 0.40°C

notice that the error bar is twice the error, which makes a lot of sense because the error (± 0.20°C) goes up and down.

Then the error bar in this measure has a magnitude of 0.40°C.

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Calculate the final concentration of BSA in the problems below using the formula
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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}

Best regards.

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Studying history enables us to develop better understanding of the world in which we live. Building knowledge and understanding of historical events and trends, especially over the past century, enables us to develop a much greater appreciation for current events today.

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