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Chapter 3: Problem 3

Digitized pictures on a disk drive: The hard disk drive on a computer holds 800 gigabytes of information. That is 800,000 megabytes. The formatting information, operating system, and applications software take up 6000 megabytes of disk space. The operator wants to store on his computer a collection of digitized pictures, each of which requires 2 megabytes of storage space. a. We think of the total amount of storage space used on the disk drive as a function of the number of pictures that are stored on the drive. Explain why this function is linear. b. Find a formula to express the total amount of storage space used on the disk drive as a linear function of the number of pictures that are stored on the drive. (Be sure to identify what the letters you use mean.) Explain in practical terms what the slope of this function is. c. Express using functional notation the total amount of storage space used on the disk drive if there are 350 pictures stored on the drive, and then calculate that value. d. After putting a number of pictures on the disk drive, the operator executes a directory command, and at the end of the list the computer displays the message \(769,000,000,000\) bytes free. This message means that there are 769,000 megabytes of storage space left on the computer. How many pictures are stored on the disk drive? How many additional pictures can be added before the disk drive is filled?

Short Answer

Expert verified
The function is linear: \( S(n) = 6000 + 2n \). For 350 pictures, storage used is 7000 MB. Currently, 12,500 pictures are stored, and 384,500 more can be added.

Step by step solution

01

Total Storage Function Explanation

The storage function is linear because the total storage used on the disk increases at a constant rate of 2 megabytes per picture. Linear functions are characterized by a constant change, or slope, for every unit increase in the independent variable (number of pictures).
02

Derive the Linear Function Formula

Let \( S(n) \) represent the total storage used in megabytes and \( n \) be the number of pictures. The function can be expressed as \( S(n) = 6000 + 2n \). Here, 6000 is the initial storage used by the operating system and applications, and 2 is the constant storage required per picture (slope).
03

Interpret the Slope

The slope of the function is 2, which means that for each additional picture stored, the disk usage increases by 2 megabytes.
04

Calculate Storage for 350 Pictures

Using the formula \( S(n) = 6000 + 2n \), substitute \( n = 350 \) to find the storage used: \( S(350) = 6000 + 2(350) = 7000 \) megabytes.
05

Determine Current and Additional Pictures

The free space available is 769,000 megabytes. Thus, the total storage used is \( 800,000 - 769,000 = 31,000 \) megabytes. Using \( S(n) = 6000 + 2n \), solve for \( n \): \( 31,000 = 6000 + 2n \), resulting in \( n = 12,500 \) pictures. The maximum storage allows for \( 800,000 = 6000 + 2n \), which gives \( n = 397,000 \) pictures. Therefore, \( 397,000 - 12,500 = 384,500 \) additional pictures can be stored.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Storage Capacity
When discussing storage capacity, it's essential to understand what it entails. Storage capacity refers to the amount of data that a storage device like a hard disk drive can hold. In the context of this exercise, we're dealing with a rapidly filling storage space as pictures are added. Each picture takes up a finite amount of space, contributing to the total storage capacity utilized. The original capacity here is 800 gigabytes, equivalent to 800,000 megabytes, which provides a large but not infinite amount of space.
  • Every digital element (such as operating systems or applications) occupies set storage.
  • The 6,000 megabytes reserved for system use is a fixed part of this capacity.
  • Additional storage is consumed as each picture (2 megabytes) is added.
Understanding this helps us manage and predict how much more data or how many pictures can be stored based on the available and used space.
Data Storage
Data storage refers to recording digital information in a storage medium, such as a hard drive. This exercise takes a practical look at storage by introducing the concept of data usage through digitized images. Each image represents a chunk of data, using up a portion of the total storage allocation on the hard drive.
  • The hard drive's 800,000 megabytes are partitioned between system needs and user data.
  • For each image of 2 megabytes, the user data increases, reducing the remaining storage capacity.
  • With 769,000 megabytes reported as free, this suggests that the data storage has reached a threshold where only the remaining space is available for additional data and images.
This exemplifies how data storage works more broadly: initial capacity reduced by consistent data additions until constraints are met.
Slope Interpretation
In the context of a linear function, the slope is a crucial component representing the rate of change between two variables鈥攊n this case, the number of pictures and the total storage used. The linear function derived in the step-by-step solution is expressed as: \[S(n) = 6000 + 2n\] Here, \(S(n)\) represents total storage in megabytes, while \(n\) is the number of pictures stored. The slope, being 2, indicates the constant increase in storage as more pictures are stored.
  • For every picture added, storage increases by 2 megabytes鈥攖his is what the slope tells us.
  • Constant slope signifies a straight line, indicating a steady, predictable increase in storage use per image.
  • This parameter simplifies planning for storage needs as it provides a predictable pattern of growth.
Understanding the slope makes it easier to manage space efficiently and foresee when upgrades or additional storage might be necessary.

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Most popular questions from this chapter

A wheelchair service ramp: The Americans with Disabilities Act (ADA) requires, among other things, that wheelchair service ramps have a slope not exceeding \(\frac{1}{12}\). a. Suppose the front steps of a building are 2 feet high. You want to make a ramp conforming to ADA standards that reaches from the ground to the top of the steps. How far away from the building is the base of the ramp? b. Another way to give specifications on a ramp is to give allowable inches of rise per foot of run. In these terms, how many inches of rise does the ADA requirement allow in 1 foot of run?

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Measuring the circumference of the Earth: Eratosthenes, who lived in Alexandria around 200 B.C., learned that at noon on the summer solstice the sun shined vertically into a well in Syene (modern Aswan, due south of Alexandria). He found that on the same day in Alexandria the sun was about 7 degrees short of being directly overhead. Since 7 degrees is about \(\frac{1}{50}\) of a full circle of 360 degrees, he concluded that the distance from Syene to Alexandria was about \(\frac{1}{50}\) of the Earth's circumference. He knew from travelers that it was a 50-day trip and that camels could travel 100 stades \(^{13}\) per day. What was Eratosthenes' measure, in stades, of the circumference of the Earth? One estimate is that the stade is \(0.104\) mile. Using this estimate, what was Eratosthenes' measurement of the circumference of the Earth in miles?

Energy cost of running: Physiologists have studied the steady-state oxygen consumption (measured per unit of mass) in a running animal as a function of its velocity (i.e., its speed). They have determined that the relationship is approximately linear, at least over an appropriate range of velocities. The table on the following page gives the velocity \(v\), in kilometers per hour, and the oxygen consumption \(E\), in milliliters of oxygen per gram per hour, for the rhea, a large, flightless South American bird. \({ }^{27}\) (For comparison, 10 kilometers per hour is about \(6.2\) miles per hour.) $$ \begin{array}{|c|c|} \hline \text { Velocity } v & \text { Oxygen consumption } E \\ \hline 2 & 1.0 \\ \hline 5 & 2.1 \\ \hline 10 & 4.0 \\ \hline 12 & 4.3 \\ \hline \end{array} $$ a. Find the equation of the regression line for \(E\) in terms of \(v\). b. The slope of the linear function giving oxygen consumption in terms of velocity is called the cost of transport for the animal, since it measures the energy required to move a unit mass by 1 unit distance. What is the cost of transport for the rhea? c. Physiologists have determined the general approximate formula \(C=8.5 \mathrm{~W}^{-0.40}\) for the cost of transport \(C\) of an animal weighing \(W\) grams. If the rhea weighs 22,000 grams, is its cost of transport from part b higher or lower than what the general formula would predict? Is the rhea a more or a less efficient runner than a typical animal its size? d. What would your equation from part a lead you to estimate for the oxygen consumption of a rhea at rest? Would you expect that estimate to be higher or lower than the actual level of oxygen consumption of a rhea at rest?

Male and female high school graduates: The table below shows the percentage of male and female high school graduates who enrolled in college within 12 months of graduation. \({ }^{31}\) $$ \begin{array}{|l|c|c|c|c|} \hline \text { Year } & 1960 & 1965 & 1970 & 1975 \\ \hline \text { Males } & 54 \% & 57.3 \% & 55.2 \% & 52.6 \% \\ \hline \text { Females } & 37.9 \% & 45.3 \% & 48.5 \% & 49 \% \\ \hline \end{array} $$ a. Find the equation of the regression line for percentage of male high school graduates entering college as a function of time. b. Find the equation of the regression line for percentage of female high school graduates entering college as a function of time. c. Assume that the regression lines you found in part a and part b represent trends in the data. If the trends persisted, when would you expect first to have seen the same percentage of female and male graduates entering college? (You may be interested to know that this actually occurred for the first time in 1980 . The percentages fluctuated but remained very close during the \(1980 \mathrm{~s}\). In the 1990 s significantly more female graduates entered college than did males. In 1992 , for example, the rate for males was \(59.6 \%\) compared with \(63.8 \%\) for females.)
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