determine the location of the last significant place value by placing a bar over the digit. (example: 1.700)…

Explanation:

Step1: Recall significant - figure rules

Non - zero digits are always significant. Zeros between non - zero digits are significant. Trailing zeros in a number with a decimal point are significant. Trailing zeros in a whole number without a decimal point may or may not be significant, but here we assume they are not significant unless otherwise indicated.

Step2: Identify last significant digit for each number

1. For 8040, assuming no decimal point, the last significant digit is 4, so it is 80(\overline{4})0.
2. For 0.0300, the last significant digit is the second 0 after the 3, so it is 0.03(\overline{0})0.
3. For 699.5, the last significant digit is 5, so it is 699.(\overline{5}).
4. For 2.000×10², the last significant digit is the last 0, so it is 2.00(\overline{0})×10².
5. For 0.90100, the last significant digit is the last 0, so it is 0.901(\overline{0})0.
6. For 90,100, assuming no decimal point, the last significant digit is 1, so it is 90,(\overline{1})00.
7. For 4.7×10⁻⁸, the last significant digit is 7, so it is 4.(\overline{7})×10⁻⁸.
8. For 10,800,000., the last significant digit is the last 0 (because of the decimal point), so it is 10,800,00(\overline{0}).
9. For 3.01×10²¹, the last significant digit is 1, so it is 3.0(\overline{1})×10²¹.
10. For 0.000410, the last significant digit is the last 0, so it is 0.00041(\overline{0}).

Answer:

1. 80(\overline{4})0
2. 0.03(\overline{0})0
3. 699.(\overline{5})
4. 2.00(\overline{0})×10²
5. 0.901(\overline{0})0
6. 90,(\overline{1})00
7. 4.(\overline{7})×10⁻⁸
8. 10,800,00(\overline{0})
9. 3.0(\overline{1})×10²¹
10. 0.00041(\overline{0})