Question

helpppppp!! The mass of a car is 1990 kg rounded to the nearest kilogram. The mass of a person is 58.7 kg rounded to 1 decimal place. Write the error interval for the combined mass, m , of the car and the person in the form a ≤ m < b

Answer 1

To find the error interval for the combined mass of the car and the person, we need to consider the possible maximum and minimum values for the masses.

For the car, since it is rounded to the nearest kilogram, the actual mass could be anywhere between 1989.5 kg and 1990.5 kg.

For the person, since it is rounded to 1 decimal place, the actual mass could be anywhere between 58.65 kg and 58.75 kg.

To find the maximum and minimum combined masses, we add the maximum possible mass of the car (1990.5 kg) to the maximum possible mass of the person (58.75 kg) and we add the minimum possible mass of the car (1989.5 kg) to the minimum possible mass of the person (58.65 kg):

Maximum combined mass = 1990.5 kg + 58.75 kg = 2049.25 kg

Minimum combined mass = 1989.5 kg + 58.65 kg = 2048.15 kg

Therefore, the error interval for the combined mass, m, of the car and the person is:

2048.15 kg ≤ m < 2049.25 kg

Answer 2

The mass of the car rounded to the nearest kilogram is 1990 kg, which has an error interval of 1989.5 kg ≤ car mass < 1990.5 kg.

The mass of the person rounded to 1 decimal place is 58.7 kg, which has an error interval of 58.65 kg ≤ person mass < 58.75 kg.

To find the error interval for the combined mass, we need to add the lower and upper bounds of the two intervals:

1989.5 kg + 58.65 kg = 2048.15 kg

1990.5 kg + 58.75 kg = 2049.25 kg

Therefore, the error interval for the combined mass, m, of the car and the person is: 2048.15 kg ≤ m < 2049.25 kg