/*! 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 18 In September 1997 , the average ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 18

In September 1997 , the average cost for 1 gallon of home heating oil in New York City was \(\$ 1.192\) per gallon. By September 2008 , it had risen to \(\$ 4.173\) per gallon. What was the percent increase in those 11 years? Round to the nearest percent.

Short Answer

Expert verified
The cost of heating oil increased by 250% over the 11 years from 1997 to 2008.

Step by step solution

01

Identify the old and new values

Here, the old value (1997 cost) is \$1.192 per gallon, and the new value (2008 cost) is \$4.173 per gallon.
02

Plug values into the percentage increase formula

We would input these values into the formula for percentage increase, which is \(\frac {new value - old value} {old value} \times 100\%\). This would be \(\frac {\$4.173 - \$1.192} {\$1.192} \times 100%\).
03

Calculate the change.

We would now perform the above calculation to determine the percentage increase. That would be \(\frac {\$2.981} {\$1.192} \times 100\%\).
04

Round to the nearest percentage.

The calculated percentage increase is approximately 250.084\%. Rounding to the nearest percent, we would get an increase of 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.

Math Problem Solving
Math problem solving is about breaking down complex problems into smaller, manageable steps. When tackling percentage increase calculations, it's essential to understand each component of the process.

In our problem, we start by identifying the old value (\(1997: \\)1.192\() and the new value (\)2008: \\(4.173\)). Recognizing and verifying these initial numbers is crucial because they form the basis of all subsequent calculations.

Once we have our values, we utilize a specific formula for percentage increase. This method ensures the problem is handled systematically and can be applied to similar problems in various contexts. Understanding and applying formulas accurately is a fundamental skill in math problem solving.
Inflation Analysis
Inflation analysis examines how prices increase over time, impacting purchasing power. In our example, the cost of heating oil rose significantly over 11 years.

This rise is a type of inflation, where essential goods and services become more expensive due to various factors. Analyzing such examples helps us understand how inflation affects everyday expenses and the broader economy.

Inflation can result from increased production costs, higher demand, or government policy changes. It's important to study these factors and their implications on household budgets and business operations.
Applied Mathematics
Applied mathematics involves using mathematical models and techniques to solve real-world problems, such as calculating percentage increases in prices.

The heating oil example is practical as it reflects how mathematical operations apply to real-life financial scenarios. By calculating the percentage increase, we can quantify how much more expensive heating oil became over time, providing essential data for budgeting and forecasting.

Using math to interpret these changes helps individuals and businesses make informed decisions based on trends and projections.
Mathematical Operations
Mathematical operations, like addition, subtraction, and division, are fundamental when calculating percentage increases.

In our exercise, we first subtract the old cost from the new cost to find the increase: $ 4.173 - 1.192 = 2.981 $ . Next, this result is divided by the old cost: $ 2.981/1.192 $ . Finally, we multiply by 100 to convert it into a percentage.

Each operation plays a critical role in reaching the solution, demonstrating how mathematical concepts are interconnected and essential for solving diverse problems.

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

Construct a pie chart that shows the following transportation-related expenses: Fuel: \(\$ 240\) ; Insurance: \(\$ 80 ;\) Public transportation: \(\$ 200\) ; Parking garage: \(\$ 120 ;\) Repairs: \(\$ 160\) .

Create a year-long budget check-off matrix to chart the following transportation related expenses: Fuel: monthly; Insurance: quarterly; Servicing: every three months; Car wash: bimonthly; Parking: semi- annually; Public transportation: monthly.

Under his household expense budget category, Mark has allocated \(\$ 60\) per month for pet food. He can purchase wet food in a can for \(\$ 1.50\) per can or dry food in a bag for \(\$ 3\) per bag. a. Determine a budget line equation for this situation. b. Graph the budget line that will depict the different combinations of cans and bags that Mark can purchase while still remaining within his budget. c. Name a combination of bags and cans that will allow him to d. Name a combination of bags and cans that will keep him under budget. e. Name a combination of bags and cans that will cause him to be over budget.

Jessica鈥檚 parents are always telling her to turn off the lights when she leaves a room. A light bulb requires 75 watts to run when it is turned on. The fi xture in Jessica鈥檚 room requires four of these bulbs. a. Jessica鈥檚 parents estimate that she leaves the lights on unnecessarily for 2.5 hours per day. How many watt-hours of electricity are used by these bulbs during 2.5 hours? b. Approximately how many kilowatt-hours of electricity are used in a year to keep these bulbs lit for 2.5 hours per day? c. At a cost of \(\$ 0.09\) per kilowatt-hour, how much money is wasted per year by keeping these lights on unnecessarily? Round to the nearest dollar. d. If five million teenagers keep lights on as Jessica does, how much is wasted in unnecessary electric expenses?

A local cable TV/Internet/phone provider charges new customers \(\$ 99\) for all three services, per month, for the first year under their \( 3\) for] \(99^{\prime \prime}\) promotion. Joanne normally pays \(\$ 54\) for monthly home phone service, \(\$ 39\) for Internet service, and \(\$ 49\) for cable television. a. What are her percent savings if she switches to the 3 for 99 plan? Round to the nearest percent. b. If, after the first year, the flat fee for all three services is \(\$ 129\) , what are her percent savings? c. Craig usually pays \(p\) dollars for phone service, \(i\) dollars for Internet service, and c dollars for cable TV service monthly. Represent savings under the 3 for 99 plan algebraically, as a percent.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

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.