Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 5

Reliability Engineering - Quality control. (a) Quality control checks find six faulty items from 200 that were tested. Express the number of faulty items as a fraction, in its simplest form, of the total number tested. (b) Improvements to the production process mean that the number of faulty items is halved. Express the number of faulty items now as a fraction of the total number tested.

Short Answer

Expert verified
Answer: The initial fraction of faulty items was 3/100, and the fraction of faulty items after improvements is 3/400.

Step by step solution

01

Find the fraction of faulty items initially

To find the fraction of faulty items initially, divide the number of faulty items by the total number of items tested. The fraction will be: 6 (faulty items) / 200 (total items tested)
02

Simplify the initial fraction

Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor. The greatest common divisor of 6 and 200 is 2. So, dividing both by 2 gives: (6/2) / (200/2) = 3/100 So, the initial fraction of faulty items is 3/100.
03

Calculate the fraction after improvements

After improvements, the number of faulty items is halved. To find the new fraction, divide the number of faulty items after improvements by the total number of items tested. The new fraction of faulty items is: (3 (initial faulty items)/2) / 200 (total items tested)
04

Simplify the fraction after improvements

Simplify the fraction as follows: (3/2) / 200 = (3/2) * (1/200) = (3/400) So, the fraction of faulty items after improvements is 3/400. The answers to the questions are: (a) The initial fraction of faulty items is 3/100. (b) The fraction of faulty items after improvements is 3/400.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Evaluate (a) \(6 \frac{2}{3} \div 4\) (b) \(10 \div 2 \frac{1}{3}\) (c) \(\left(\frac{1}{2}+\frac{1}{3}\right) \div\left(\frac{2}{3}+\frac{1}{5}\right)\) (d) \(\left(6-2 \frac{1}{3}\right) \times\left(4 \frac{1}{2}-1 \frac{3}{4}\right)\) (e) \(\frac{2 \frac{1}{2}+1 \frac{1}{3}}{6 \frac{2}{3}-2 \frac{1}{4}}\)

Express the following mixed fractions as improper fractions: (a) \(1 \frac{1}{2}\) (b) \(2 \frac{1}{3}\) (c) \(3 \frac{1}{4}\) (d) \(3 \frac{2}{5}\) (e) \(-10 \frac{2}{5}\)

Calculate (a) \(\frac{6}{7} \times \frac{14}{27}\) (b) \(\frac{7}{10} \times \frac{4}{5} \times \frac{30}{49}\) (c) \(\frac{8}{9} \times \frac{18}{25}\) (d) \(\left(-\frac{4}{5}\right) \times\left(-\frac{3}{4}\right)\) (e) \(\frac{16}{21} \times\left(-\frac{3}{4}\right)\)

Calculate (a) \(\frac{7}{9} \div \frac{2}{3}\) (b) \(\frac{8}{15} \div \frac{4}{5}\) (c) \(\frac{9}{10} \div \frac{9}{20}\) (d) \(\frac{6}{11} \div \frac{7}{12}\) (e) \(\frac{12}{13} \div \frac{6}{7}\)

Dynamics - Pressure in a vessel. The pressure in a vessel is \(30 \mathrm{~N} \mathrm{~cm}^{-2}\). If the pressure is reduced by \(\frac{7}{100}\) of its original value, calculate (a) the decrease in pressure, (b) the resulting pressure in the vessel.
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Particle Model of Matter

Read Explanation

Physics of Motion

Read Explanation

Dynamics

Read Explanation

Nuclear Physics

Read Explanation

Atoms and Radioactivity

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.

Sign-up for free