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To determine which additional set of values could be included in the given table, we need to identify if the relationship between depth and pressure is proportional. This means that the pressure should increase at a constant rate with respect to the depth.
Here are the given data points representing depth ([tex]$x$[/tex]) and pressure ([tex]$y$[/tex]):
[tex]\[ (0, 0), (15, 3), (30, 6) \][/tex]
First, we will calculate the constant rate (ratio) of pressure change per meter of depth. We can do this by taking any of the given data points (other than the origin) and dividing the pressure by the depth.
Using the points [tex]\((15, 3)\)[/tex]:
[tex]\[ \text{Ratio} = \frac{y}{x} = \frac{3 \text{ atm}}{15 \text{ m}} = 0.2 \text{ atm/m} \][/tex]
We now have determined that for every meter of depth, the pressure increases by [tex]\(0.2\)[/tex] atmospheres. This relationship should hold true for all valid additional points.
Next, we’ll test each of the provided additional sets of values to see if they follow this same proportional relationship:
1. For the set [tex]\((10, 2)\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{2 \text{ atm}}{10 \text{ m}} = 0.2 \text{ atm/m} \][/tex]
Since the ratio is [tex]\(0.2\)[/tex], this set could be included in the table.
2. For the set [tex]\((40, 9)\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{9 \text{ atm}}{40 \text{ m}} = 0.225 \text{ atm/m} \][/tex]
Since the ratio is not [tex]\(0.2\)[/tex], this set cannot be included.
3. For the set [tex]\((50, 38)\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{38 \text{ atm}}{50 \text{ m}} = 0.76 \text{ atm/m} \][/tex]
Since the ratio is not [tex]\(0.2\)[/tex], this set cannot be included.
4. For the set [tex]\((100, 76)\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{76 \text{ atm}}{100 \text{ m}} = 0.76 \text{ atm/m} \][/tex]
Since the ratio is not [tex]\(0.2\)[/tex], this set cannot be included.
Considering the proportional relationship requirement and the calculated ratios, the only additional set of values that could be included in the table is [tex]\((10, 2)\)[/tex].
Here are the given data points representing depth ([tex]$x$[/tex]) and pressure ([tex]$y$[/tex]):
[tex]\[ (0, 0), (15, 3), (30, 6) \][/tex]
First, we will calculate the constant rate (ratio) of pressure change per meter of depth. We can do this by taking any of the given data points (other than the origin) and dividing the pressure by the depth.
Using the points [tex]\((15, 3)\)[/tex]:
[tex]\[ \text{Ratio} = \frac{y}{x} = \frac{3 \text{ atm}}{15 \text{ m}} = 0.2 \text{ atm/m} \][/tex]
We now have determined that for every meter of depth, the pressure increases by [tex]\(0.2\)[/tex] atmospheres. This relationship should hold true for all valid additional points.
Next, we’ll test each of the provided additional sets of values to see if they follow this same proportional relationship:
1. For the set [tex]\((10, 2)\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{2 \text{ atm}}{10 \text{ m}} = 0.2 \text{ atm/m} \][/tex]
Since the ratio is [tex]\(0.2\)[/tex], this set could be included in the table.
2. For the set [tex]\((40, 9)\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{9 \text{ atm}}{40 \text{ m}} = 0.225 \text{ atm/m} \][/tex]
Since the ratio is not [tex]\(0.2\)[/tex], this set cannot be included.
3. For the set [tex]\((50, 38)\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{38 \text{ atm}}{50 \text{ m}} = 0.76 \text{ atm/m} \][/tex]
Since the ratio is not [tex]\(0.2\)[/tex], this set cannot be included.
4. For the set [tex]\((100, 76)\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{76 \text{ atm}}{100 \text{ m}} = 0.76 \text{ atm/m} \][/tex]
Since the ratio is not [tex]\(0.2\)[/tex], this set cannot be included.
Considering the proportional relationship requirement and the calculated ratios, the only additional set of values that could be included in the table is [tex]\((10, 2)\)[/tex].
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