SAT Question 2a59eb45: Distributions, Center & Spread answer and explanation with Desmos solution

The mean of a data set is computed by dividing the sum of the values in the data set by the number of values in the data set. It's given that data set A consists of the heights of $75$ buildings and has a mean of $32$ meters. This can be represented by the equation $\frac{x}{75}=32$, where $x$ represents the sum of the heights of the buildings, in meters, in data set A. Multiplying both sides of this equation by $75$ yields $x=75\left(32\right)$, or $x=2,400$ meters. Therefore, the sum of the heights of the buildings in data set A is $2,400$ meters. It's also given that data set B consists of the heights of $50$ buildings and has a mean of $62$ meters. This can be represented by the equation $\frac{y}{50}=62$, where $y$ represents the sum of the heights of the buildings, in meters, in data set B. Multiplying both sides of this equation by $50$ yields $y=50\left(62\right)$, or $y=3,100$ meters. Therefore, the sum of the heights of the buildings in data set B is $3,100$ meters. Since it's given that data set C consists of the heights of the $125$ buildings from data sets A and B, it follows that the mean of data set C is the sum of the heights of the buildings, in meters, in data sets A and B divided by the number of buildings represented in data sets A and B, or $\frac{2,400+3,100}{125}$, which is equivalent to $44$ meters. Therefore, the mean, in meters, of data set C is $44$.