A farmer recorded the amount of nitrogen fertilizer applied, (in tens of kilograms per hectare), and the corn yield, (in kilograms per hectare). The results are shown in the scatterplot, along with a line of best fit. The farmer plans to apply fertilizer to a -hectare field and wants a predicted yield of kilograms per hectare. The fertilizer is sold in -kilogram bags. Which choice is the minimum number of bags the farmer should buy, according to the line of best fit?

Solution available with step-by-step explanation

SAT Two-Variable Data Question 27 | Free SAT Prep | Aniko

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Question 27·Hard·Two-Variable Data: Models and Scatterplots

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A farmer recorded the amount of nitrogen fertilizer applied, $x$ (in tens of kilograms per hectare), and the corn yield, $y$ (in kilograms per hectare). The results are shown in the scatterplot, along with a line of best fit.

The farmer plans to apply fertilizer to a $6.5$ -hectare field and wants a predicted yield of $465$ kilograms per hectare. The fertilizer is sold in $25$ -kilogram bags.

Which choice is the minimum number of bags the farmer should buy, according to the line of best fit?

For best-fit-line prediction problems, choose two clean points on the line (ideally grid intersections), compute $m$ and write $y=mx+b$ , then substitute the target $y$ and solve for $x$ . Finally, handle all unit conversions from the axes (like “tens of”) and apply any real-world scaling (area, packages) at the end, rounding up when the context requires a minimum whole number.

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Watch the units on the $x$ -axis

The $x$ -values are in tens of kilograms per hectare, so you’ll need to multiply by $10$ after solving for $x$ .

Use two clear points on the best-fit line

Pick two grid-intersection points on the line to find the slope and write an equation $y=mx+b$ .

After finding kg/ha, scale to the whole field

Once you have fertilizer in kg/ha, multiply by $6.5$ hectares to get total kilograms.

Bags must be whole numbers

Divide total kilograms by $25$ kg per bag. If you don’t get an integer, you need to round up to ensure there is enough fertilizer.

Desmos Guide

Enter the line through two points

In Desmos, define the two points on the line: $P=(2,210)$ and $Q=(10,390)$ . Then enter the line through them, for example:

y-210=\frac{390-210}{10-2}(x-2).

Find the $x$ -value when $y=465$

Enter $y=465$ and find the intersection with the best-fit line. The intersection’s $x$ -coordinate is in tens of kg/ha.

Convert and scale to bags

Multiply that $x$ -value by $10$ to get kg/ha, multiply by $6.5$ to get total kg, then divide by $25$ and round up to a whole number of bags.

Step-by-step Explanation

Write an equation for the line of best fit (graph units)

From the graph, the best-fit line passes through $(2,210)$ and $(10,390)$ .

Slope:

m=\frac{390-210}{10-2}=\frac{180}{8}=\frac{45}{2}

Using $(2,210)$ in point-slope form:

y-210=\frac{45}{2}(x-2)

So the line can be written as

y=\frac{45}{2}x+165.

Find the fertilizer amount (in tens of kg/ha) that predicts $y=465$

Substitute $y=465$ :

465=\frac{45}{2}x+165

300=\frac{45}{2}x

x=300\cdot\frac{2}{45}=\frac{40}{3}.

This $x$ is in tens of kilograms per hectare.

Convert to kilograms per hectare and scale to a $6.5$ -hectare field

Convert from tens of kg/ha to kg/ha:

\left(\frac{40}{3}\right)\cdot 10=\frac{400}{3}\text{ kg/ha}.

Total fertilizer for $6.5=\frac{13}{2}$ hectares:

\frac{400}{3}\cdot\frac{13}{2}=\frac{2600}{3}\text{ kg}.

Convert kilograms to bags and round up

Each bag is $25$ kg, so the number of bags needed is

\frac{\frac{2600}{3}}{25}=\frac{2600}{75}=\frac{104}{3}=34\frac{2}{3}.

Because the farmer must buy a whole number of bags, round up to the next integer.

Select the matching choice

The minimum number of bags is $35$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation