/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 (Modeling) Solve each problem. S... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 100

(Modeling) Solve each problem. See Example 11 . Employee Training A person learning certain skills involving repetition tends to learn quickly at first. Then learning tapers off and skill acquisition approaches some upper limit. Suppose the number of symbols per minute that a person using a keyboard can type is given by $$ f(t)=250-120(2.8)^{-0.5 t} $$ where \(t\) is the number of months the operator has been in training. Find each value. (a) \(f(2)\) (b) \(f(4)\) (c) \(f(10)\) (d) What happens to the number of symbols per minute after several months of training?

Short Answer

Expert verified
f(2) 鈮� 207.15, f(4) 鈮� 234.69, f(10) 鈮� 249.66. After several months, the number of symbols per minute approaches 250.

Step by step solution

01

Understand the Function

The function given is: \[ f(t) = 250 - 120(2.8)^{-0.5t} \]This function represents the number of symbols per minute (f) that a person can type after t months of training.
02

Evaluate for t = 2

Substitute t = 2 into the function: \[ f(2) = 250 - 120(2.8)^{-0.5(2)} \]Simplify the exponent first: \[ -0.5 \times 2 = -1 \]So, \[ f(2) = 250 - 120(2.8)^{-1} \]Calculate \[ 2.8^{-1} = \frac{1}{2.8} \approx 0.3571 \]Multiply: \[ 120 \times 0.3571 \approx 42.85 \]Finally, \[ f(2) = 250 - 42.85 \approx 207.15 \]
03

Evaluate for t = 4

Substitute t = 4 into the function: \[ f(4) = 250 - 120(2.8)^{-0.5(4)} \]Simplify the exponent first: \[ -0.5 \times 4 = -2 \]So, \[ f(4) = 250 - 120(2.8)^{-2} \]Calculate \[ 2.8^{-2} = \left(\frac{1}{2.8}\right)^{2} \approx 0.1276 \]Multiply: \[ 120 \times 0.1276 \approx 15.31 \]Finally, \[ f(4) = 250 - 15.31 \approx 234.69 \]
04

Evaluate for t = 10

Substitute t = 10 into the function: \[ f(10) = 250 - 120(2.8)^{-0.5(10)} \]Simplify the exponent first: \[ -0.5 \times 10 = -5 \]So, \[ f(10) = 250 - 120(2.8)^{-5} \]Calculate \[ 2.8^{-5} = \left(\frac{1}{2.8}\right)^{5} \approx 0.00280 \]Multiply: \[ 120 \times 0.00280 \approx 0.34 \]Finally, \[ f(10) = 250 - 0.34 \approx 249.66 \]
05

Analyze Long-term Behavior

As \( t \) increases, the term \( (2.8)^{-0.5t} \) approaches 0 because the exponent becomes more negative. Thus, the expression \( 250 - 120(2.8)^{-0.5t} \) approaches 250, indicating that after several months of training, the number of symbols per minute approaches 250.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Exponential Functions
The key components of exponential functions:
Long-term Behavior
Therefore:
Skill Acquisition
The exercise function \( f(t) = 250 - 120(2.8)^{-0.5t} \) models this behavior in typing skills. It shows fast initial progress and gradual leveling off. Here鈥檚 a breakdown:

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At the World Championship races held at Rome's Olympic Stadium in \(1987,\) American sprinter Carl Lewis ran the 100 -m race in 9.86 sec. His speed in meters per second after \(t\) seconds is closely modeled by the function $$f(t)=11.65\left(1-e^{-t / 1.27}\right)$$ (Source: Banks, Robert B., Towing Icebergs, Falling Dominoes, and Other Adventures in Applied Mathematics, Princeton University Press.) (a) How fast was he running as he crossed the finish line? (b) After how many seconds was he running at the rate of \(10 \mathrm{m}\) per sec?

Population Decline A midwestern city finds its residents moving to the suburbs. Its population is declining according to the function defined by $$ P(t)=P_{0} e^{-0.04 t} $$ where \(t\) is time measured in years and \(P_{0}\) is the population at time \(t=0 .\) Assume that \(P_{0}=1,000,000\) (a) Find the population at time \(t=1\) (b) Estimate the time it will take for the population to decline to \(750,000\). (c) How long will it take for the population to decline to half the initial number?

(a) Explain why a polynomial function of even degree cannot have an inverse. (b) Explain why a polynomial function of odd degree may not be one-to-one.

Deer Population The exponential growth of the deer population in Massachusetts can be calculated using the model $$ f(x)=50,000(1+0.06)^{x} $$ where \(50,000\) is the initial deer population and 0.06 is the rate of growth. \(f(x)\) is the total population after \(x\) years have passed. (a) Predict the total population after 4 yr. (b) If the initial population was \(30,000\) and the growth rate was \(0.12,\) approximately how many deer would be present after 3 yr? (c) How many additional deer can we expect in 5 yr if the initial population is \(45,000\) and the current growth rate is \(0.08 ?\) (IMAGE CANT COPY)

Refer to the formulas for compound interest. $$A=P\left(1+\frac{r}{n}\right)^{t n} \text { and } A=P e^{r t} \quad \text { (Section 4.2) }$$ At what interest rate, to the nearest hundredth of a percent, will 16,000 dollars grow to 20,000 dollars if invested for 5.25 yr and interest is compounded quarterly?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.