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Difficulty: HardInterpreting Linear Relationships in Context

A deep space communications satellite transmits a telemetry data file to a ground station on Earth. The remaining size of the file to be received at the ground station, SS, in megabytes (MB), can be modeled by the equation S=850−4.5(t−18)S = 850 - 4.5(t - 18), where tt is the number of seconds since the satellite initiated its transmission sequence, and t≥18t \ge 18. Which of the following is the best interpretation of the number 1818 in this context?

  1. A
    The rate, in megabytes per second, at which the data is received by the ground station.
  2. B
    The size, in megabytes, of the telemetry file when the ground station begins receiving it.
  3. The number of seconds after the transmission sequence is initiated before the ground station begins receiving the file.Answer
  4. D
    The total number of seconds needed to fully transmit the telemetry file.

Answer

The number of seconds after the transmission sequence is initiated before the ground station begins receiving the file.
The model S=850−4.5(t−18)S = 850 - 4.5(t - 18) is valid for t≥18t \ge 18. Substituting t=18t = 18 into the equation yields S=850S = 850 megabytes, which is the total size of the file before any data has been received. As tt increases beyond 1818, the remaining file size decreases at a rate of 4.54.5 megabytes per second. Therefore, the first 1818 seconds after the transmission sequence is initiated represent the time delay before the ground station starts receiving the file.

Step-by-Step Solution

1
Analyze the structure of the linear equation S=850−4.5(t−18)S = 850 - 4.5(t - 18) in context.
The variable SS represents the remaining file size in megabytes, and tt represents the time in seconds since the sequence was initiated. The model is defined only for t≥18t \ge 18.
Understanding the variables and constraints is the first step to interpreting the components of the equation.
2
Evaluate the equation at the boundary value t=18t = 18.
When t=18t = 18, S=850−4.5(18−18)=850S = 850 - 4.5(18 - 18) = 850 megabytes.
This determines the starting state of the data reception modeled by the equation.
3
Analyze how SS changes as tt increases beyond 1818.
For every second tt increases beyond 1818, SS decreases by 4.54.5 megabytes.
This confirms that the transmission starts at t=18t = 18 seconds and proceeds at a rate of 4.54.5 megabytes per second, meaning the first 1818 seconds represent the delay before data reception starts.

Key Concept

Interpreting Linear Relationships in Context
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Interpreting Linear Relationships in Context

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