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In Problems 17—24 the strategy is to first find a general solution of the dif- ferential equation, then solve for the constants to find a solution satisfying the initial conditions. Problems 17—22 are well suited to the use of undetermined co- efficients, while Problems 23 and 24 can be solved fairly directly using variation of parameters.
This is a constant coefficient second-order homogeneous differential equa- tion for Y t , which we know how to solve. We obtain the solution in all cases by solving this linear constant coefficient second-order equation.
We know how to solve this problem. However, also transform the initial conditions: The new twist here is that the entire initial value problem including initial conditions was transformed in terms of t and solved for Y t , then this solution Y t in terms of t was transformed back to the solution y x in terms of x.
These solutions are consistent with the observation that, upon division by 4, the differential equation is an Euler equation. Sums of pairs of dice, Problem 1, Section 2. The standard deviation is If we roll two dice, there are thirty-six possible outcomes. The sums of the numbers that can come up on the two dice are listed in Table 1.
The table gives all of the values that X o can take on, over all outcomes o of the experiment. Each value is listed as often as it occurs as a value of X.
This is interpreted to mean that, on average, we expect to come up with a seven if we roll two dice. This is a reasonable expectation in view of the fact that there are more ways to roll 7 than any other sum with two dice.
Flip four coins, with sixteen possible outcomes. If o is an outcome, X o can have only two values, namely 1 if two, three, or four tails are in o, or 3 otherwise one tail or no tails in o. The values assumed by X are 0, 1, 2, 3, 4.
These are the probabilities of the values of the random variable X. Outcomes of rolling two dice, Problem 4, Section 2. The outcomes of two rolls of the dice are displayed in Table 2. There are 52 C2 ways to do this, disregarding order.
If o is an outcome in which both cards are numbered, then X o equals the sum of the numbers on the cards. Section 3 The Binomial and Poisson Distributions 1. Now the outcomes have the appearance x, y, T , where x and y can independently be 2, 4 or 6. The fact that these are letters of the alphabet that we are arranging in order is irrelevant. The issue is that there are nine distinct objects. The number of arrangements is 9! There are 26 letters in the English slphabet. The problem is one of deter- mining the number of ways of choosing 17 objects from 26 objects, with order taken into account.
The total number of codes is 99 , or 3. These 7 symbols have 7! The number of ways of doing this is 5! For n! There are 12! There are 10 letters from a through l, inclusive.
There are 7! We want to pick 7 objects from 25, taking order into account. The number of ways to do this is 25!
The number of ballots is 16! Because order is important, the number of possibilities is 22! There are 19! This is reasonable from a common sense point of view, since, with 20 number to choose from, we would expect 5 percent to begin with any particular one of the numbers.
The answer is that 5 percent of the choices ends in 9. There are 18 such choices. Bookmark it to easily review again before an exam. The best part? As a Chegg Study subscriber, you can view available interactive solutions manuals for each of your classes for one low monthly price.
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