Problem Set. Three crates of Davao pomelos arrived at your warehouse. The crates are of equal sizes and properly labelled, except one:  Crate A is filled to the brim and weighs 11.25 kg; its label indicates a quantity of 20.0 pcs.  Crate B is only half-filled and weighs 6.25 kg; its label indicates a quantity of only 10.0 pcs.  Crate C weighs 9.75 kg, but has its label ripped off during transport. You need to reprint a new label and indicate the quantity of pomelos inside Crate C without actually opening the box.  What is the average weight of a Davao pomelo?  What is the tare weight of an empty crate?  How many pomelos are there in Crate C?  Can you predict how much a crate containing 30.0 pomelos would weigh? (Complete this problem by plotting a graph of weight in the y-axis vs. quantity of pomelos in the x-axis, and writing the linear equation y = mx + b. Your graph doesn’t need to look pretty)

To solve this problem, we need to follow these steps:

1. Calculate the average weight of a Davao pomelo: We know that Crate A has 20 pomelos and weighs 11.25 kg. So, the average weight of a pomelo is 11.25 kg / 20 = 0.5625 kg/pomelo.

2. Calculate the tare weight of an empty crate: We know that Crate B is half-filled and weighs 6.25 kg with 10 pomelos. So, the weight of the pomelos in Crate B is 10 * 0.5625 kg/pomelo = 5.625 kg. Therefore, the tare weight of the crate is 6.25 kg - 5.625 kg = 0.625 kg.

3. Calculate the number of pomelos in Crate C: We know that Crate C weighs 9.75 kg. Subtracting the tare weight, we get the weight of the pomelos in Crate C as 9.75 kg - 0.625 kg = 9.125 kg. So, the number of pomelos in Crate C is 9.125 kg / 0.5625 kg/pomelo = 16.2. Since we can't have a fraction of a pomelo, we round this to 16 pomelos.

4. Predict the weight of a crate containing 30 pomelos: The weight of the pomelos would be 30 * 0.5625 kg/pomelo = 16.875 kg. Adding the tare weight of the crate, the total weight would be 16.875 kg + 0.625 kg = 17.5 kg.

5. Plotting the graph and writing the linear equation: The linear equation would be y = mx + b, where y is the total weight of the crate, m is the weight of a pomelo (0.5625 kg/pomelo), x is the number of pomelos, and b is the tare weight of the crate (0.625 kg). So, the equation is y = 0.5625x + 0.625.

To solve this problem, we need to follow these steps:

1. Calculate the average weight of a Davao pomelo: We know that Crate A has 20 pomelos and weighs 11.25 kg. So, the average weight of a pomelo is 11.25 kg / 20 = 0.5625 kg/pomelo.

2. Calculate the tare weight of an empty crate: We know that Crate B is half-filled and weighs 6.25 kg with 10 pomelos. So, the weight of the pomelos in Crate B is 10 * 0.5625 kg/pomelo = 5.625 kg. Therefore, the tare weight of the crate is 6.25 kg - 5.625 kg = 0.625 kg.

3. Calculate the number of pomelos in Crate C: We know that Crate C weighs 9.75 kg. Subtracting the tare weight, we get the weight of the pomelos in Crate C as 9.75 kg - 0.625 kg = 9.125 kg. So, the number of pomelos in Crate C is 9.125 kg / 0.5625 kg/pomelo = 16.2. Since we can't have a fraction of a pomelo, we round this to 16 pomelos.

4. Predict the weight of a crate containing 30 pomelos: The weight of the pomelos would be 30 * 0.5625 kg/pomelo = 16.875 kg. Adding the tare weight of the crate, the total weight would be 16.875 kg + 0.625 kg = 17.5 kg.

5. Plotting the graph and writing the linear equation: The weight (y) is a function of the quantity of pomelos (x). The slope (m) of the line is the weight of a single pomelo, and the y-intercept (b) is the tare weight of the crate. So, the equation is y = 0.5625x + 0.625.