Domanda del libro di testo
Using different substitutions
Show that the integral
∫((x² - 1)(x + 1))^(-2/3) dx
can be evaluated with any of the following substitutions.
a. u = 1/(x + 1)
What is the value of the integral?
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Ch. 8 - Techniques of Integration
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 8, Problema 8.9.42a
Lifetime of a tire Assume the random variable L in Example 2f is normally distributed with mean μ = 22,000 miles and σ = 4,000 miles.
a. In a batch of 4000 tires, how many can be expected to last for at least 18,000 miles?
Guida verificata passo dopo passo1
Identify the given information: the lifetime L of tires is normally distributed with mean \( \mu = 22000 \) miles and standard deviation \( \sigma = 4000 \) miles. The batch size is 4000 tires.
We want to find the number of tires expected to last at least 18,000 miles. This means we need to find \( P(L \geq 18000) \), the probability that a tire lasts 18,000 miles or more.
Convert the lifetime value 18,000 miles to a standard normal variable \( Z \) using the formula: \[ Z = \frac{X - \mu}{\sigma} = \frac{18000 - 22000}{4000} \]. This standardizes the value to the standard normal distribution.
Use the standard normal distribution table or a calculator to find \( P(Z \geq z) \), where \( z \) is the value calculated in the previous step. This gives the probability that a tire lasts at least 18,000 miles.
Multiply the probability \( P(L \geq 18000) \) by the total number of tires (4000) to find the expected number of tires lasting at least 18,000 miles: \[ \text{Expected number} = 4000 \times P(L \geq 18000) \].
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Normal Distribution
The normal distribution is a continuous probability distribution characterized by its bell-shaped curve, defined by the mean (μ) and standard deviation (σ). It models many natural phenomena, including tire lifetimes, and allows calculation of probabilities for values within certain ranges.
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06:11
Limits of Rational Functions: Denominator = 0
Standardization (Z-Score)
Standardization converts a normal random variable to a standard normal variable (Z) with mean 0 and standard deviation 1. This is done using the formula Z = (X - μ) / σ, enabling the use of standard normal tables to find probabilities associated with specific values.
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Percorso guidato
3:40Circles in Standard Form Example 1
Expected Value in Probability
The expected value represents the average or mean outcome in a probabilistic scenario. When dealing with large batches, multiplying the probability of an event by the total number of trials estimates how many items meet a certain criterion, such as tires lasting at least 18,000 miles.
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Percorso guidato
06:37Average Value of a Function
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