Question

round 785.85 to the indicated number of significant figures:

1. 2 significant figures
2. 3 significant figures
3. 4 significant figures
4. 1 significant figure

complete the following calculations, expressing your answer to the correct number of significant digits.

1. 24.57 + 5.678 + 5.3

49..0012 ÷ 5.9

1. 59.34 × 5
2. 8.895 - 6.22

use dimensional analysis to solve the following problems:

1. how many microliters are in 0.00528 liters?
2. what portion of a year is 3.97 × 10¹⁵ nanoseconds?
3. convert 0.75 hm³ to mm³.

Explanation:

Step1: Solve problem 48

Add the numbers: $24.57+5.678 + 5.3=35.548$. Rounding to the correct number of significant - digits (the least number of decimal places among the numbers being added is 1), the answer is $35.5$.

Step2: Solve problem 49

Calculate $0.0012\div5.9=\frac{0.0012}{5.9}\approx0.000203$. Rounding to 2 significant digits (since 0.0012 has 2 significant digits), the answer is $2.0\times10^{- 4}$.

Step3: Solve problem 50

Multiply $59.34\times5 = 296.7$. Rounding to 2 significant digits (since 5 has 1 significant digit, but we usually consider the non - exact number's significant digits, and 59.34 has 4 significant digits, and in multiplication we go by the least number of significant digits among non - exact numbers in a more complex sense, here we can consider 5 as having at least 1 significant digit and round 296.7 to 300).

Step4: Solve problem 51

Subtract $8.895−6.22 = 2.675$. Rounding to 3 significant digits (since 6.22 has 3 significant digits), the answer is $2.68$.

Step5: Solve problem 52

Use the conversion factor $1\ liter = 10^{6}\ \mu L$. So, for $0.00528\ L$, we have $0.00528\times10^{6}\ \mu L=5280\ \mu L$.

Step6: Solve problem 53

First, convert $3.97\times 10^{15}\ ns$ to seconds. Since $1\ ns=10^{-9}\ s$, then $3.97\times 10^{15}\ ns = 3.97\times10^{15}\times10^{-9}\ s=3.97\times10^{6}\ s$.
A year has approximately $365\times24\times3600\ s=31536000\ s$.
The portion of a year is $\frac{3.97\times 10^{6}}{31536000}\approx0.126$.

Step7: Solve problem 54

Use the conversion factors: $1\ hm = 100\ m$ and $1\ m = 1000\ mm$. So $1\ hm=10^{5}\ mm$, and $1\ hm^{3}=(10^{5})^{3}\ mm^{3}=10^{15}\ mm^{3}$.
Then $0.75\ hm^{3}=0.75\times10^{15}\ mm^{3}=7.5\times 10^{14}\ mm^{3}$.

Answer:

1. $35.5$
2. $2.0\times10^{-4}$
3. $300$
4. $2.68$
5. $5280\ \mu L$
6. $0.126$
7. $7.5\times 10^{14}\ mm^{3}$