Question

1. paint coverage for one coat of paint is typically 10.7 m²/l. how much paint (in qt) is required to paint 42.8 yd² of walls? assume one coat of paint. make sure to express your answer to the correct number of significant figures. (1.057 qt = 1 l; 1 m = 1.094 yd) (18 pts)

Explanation:

Step1: Convert square yards to square meters

First, we know that \(1\space\text{yd} = 1.094\space\text{m}\), so for area, \((1\space\text{yd})^2=(1.094\space\text{m})^2\). Then \(1\space\text{yd}^2=(1.094)^2\space\text{m}^2\approx1.197\space\text{m}^2\).
The area of the wall is \(42.8\space\text{yd}^2\), so in square meters, it is \(42.8\times1.197\space\text{m}^2\approx 42.8\times1.2 = 51.36\space\text{m}^2\) (we can also calculate it precisely: \(42.8\times1.197 = 42.8\times(1 + 0.197)=42.8+42.8\times0.197=42.8 + 8.4316 = 51.2316\space\text{m}^2\)).

Step2: Find the volume of paint in liters

The paint coverage is \(10.7\space\text{m}^2/\text{L}\). So the volume of paint in liters \(V=\frac{\text{Area}}{\text{Coverage}}=\frac{51.2316\space\text{m}^2}{10.7\space\text{m}^2/\text{L}}\approx4.7879\space\text{L}\).

Step3: Convert liters to quarts

We know that \(1\space\text{L}=1.057\space\text{qt}\), so the volume in quarts is \(V = 4.7879\space\text{L}\times1.057\space\text{qt/L}\approx4.7879\times1.057\approx5.06\space\text{qt}\).

Now, let's check the significant figures. The given values: \(42.8\) (3 sig figs), \(10.7\) (3 sig figs), \(1.094\) (4 sig figs), \(1.057\) (4 sig figs). When multiplying and dividing, the result should have the same number of significant figures as the least precise measurement. Here, \(42.8\) and \(10.7\) have 3 sig figs, so our final answer should have 3 sig figs.

Rounding \(5.06\) to 3 significant figures, we get \(5.06\approx5.06\) (wait, actually, let's recalculate more precisely:

First, convert \(42.8\space\text{yd}^2\) to \(\text{m}^2\):

\(1\space\text{yd}=1.094\space\text{m}\), so \(1\space\text{yd}^2=(1.094)^2\space\text{m}^2 = 1.196836\space\text{m}^2\)

\(42.8\space\text{yd}^2\times1.196836\space\text{m}^2/\text{yd}^2=42.8\times1.196836 = 42.8\times1 + 42.8\times0.196836=42.8+8.3246808 = 51.1246808\space\text{m}^2\)

Then, volume in liters: \(51.1246808\space\text{m}^2\div10.7\space\text{m}^2/\text{L}=51.1246808\div10.7\approx4.777998\space\text{L}\)

Then, convert to quarts: \(4.777998\space\text{L}\times1.057\space\text{qt/L}=4.777998\times1.057\)

\(4.777998\times1.057 = 4.777998+4.777998\times0.057=4.777998 + 0.272345886=5.050343886\space\text{qt}\)

Now, \(5.050343886\) with 3 significant figures is \(5.05\) (wait, no: the first non - zero digit is 5, then 0, then 5. Wait, 5.05034... So the first three significant figures are 5, 0, 5? Wait, no: 5.05034, the significant figures are 5, 0, 5, 0,... Wait, no, 5.05034: the first significant figure is 5, second is 0 (because it's between two non - zero digits), third is 5. So 5.05 when rounded to 3 significant figures? Wait, no, 5.05034, the third significant figure is 5, the next digit is 0, which is less than 5, so we keep it as 5.05? Wait, no, let's count:

Number: 5.05034

Significant figures:

1st: 5

2nd: 0 (because it's between 5 and 5, so it's significant)

3rd: 5

4th: 0

So when rounding to 3 significant figures, we look at the 4th digit, which is 0, so we don't round up. So it's 5.05? But wait, our calculation steps:

Wait, maybe I made a mistake in the conversion steps. Let's do it step by step with exact conversions:

1. Convert \(A = 42.8\space\text{yd}^2\) to \(\text{m}^2\):

Since \(1\space\text{yd}=1.0936133\space\text{m}\) (more precise value), so \(1\space\text{yd}^2=(1.0936133)^2\space\text{m}^2 = 1.196033\space\text{m}^2\)

\(A = 42.8\space\text{yd}^2\times1.196033\space\text{m}^2/\text{yd}^2=42.8\times1.196033 = 42.8\times1+42.8\times0.196033=42.8 + 8.3902124 = 51.1902124\space\text{m}^2\)

2. Volume in liters: \(V_L=\frac{A}{\text{coverage}}=\frac{51.1902124\space\text{m}^2}{10.7\space\text{m}^2/\text{L}}=51.1902124\div10.7\approx4.784132\space\text{L}\)

3. Volume in quarts: \(V_{qt}=V_L\times1.056688\space\text{qt/L}\) (more precise conversion factor: \(1\space\text{L}=1.056688\space\text{qt}\))

\(V_{qt}=4.784132\times1.056688\approx4.784132\times1.056688\)

Calculate \(4\times1.056688 = 4.226752\)

\(0.784132\times1.056688\approx0.784132\times1+0.784132\times0.056688\approx0.784132 + 0.04445\approx0.828582\)

Total \(V_{qt}\approx4.226752+0.828582 = 5.055334\space\text{qt}\)

Now, the number of significant figures: the area \(42.8\) has 3, coverage \(10.7\) has 3, so the result should have 3 significant figures. \(5.055334\) rounded to 3 significant figures is \(5.06\) (wait, because the fourth digit is 5, so we round up the third digit: 5.055... the third digit is 5, the next digit is 5, so we round up 5 to 6? Wait, no:

The number is 5.055334

Significant figures:

1: 5

2: 0

3: 5

4: 5

So when rounding to 3 significant figures, we look at the digit after the third (the 4th digit, which is 5). So we round the third digit up: 5.05 + 0.01 = 5.06

Yes, because the rule is: if the digit to the right of the last significant figure is 5 or more, we round up the last significant figure.

So the final answer is approximately \(5.06\space\text{qt}\) (with 3 significant figures).

Answer:

\(\boxed{5.06}\) (or if we consider more precise calculation, maybe \(5.06\) with 3 significant figures)