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Chapter 7: Q. 25 (page 639)

Simplify the quotients in Exercises 21–28 without using a calculator.
$\frac{4!}{(2!)!}$

Short Answer

Expert verified

The value is $12 .$

Step by step solution

Step 1. Given information.

The given expression is $\frac{4!}{(2!)!}$ .

Step 2. Value of the expression.

Now, we know,

$n! = n (n - 1)! T h e r e f o r e, \frac{4!}{(2!)!} = \frac{4!}{(2 \times 1)!} = \frac{4!}{2!} = \frac{(4) (3) 2!}{2!} = 12$

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Most popular questions from this chapter

Let $\sum_{k = 1}^{\infty} a_{k} b e a c o n v e r g e n t s e r i e s a n d \sum_{k = 1}^{\infty} b_{k} b e a d i v e r g e n t s e r i e s .$ Prove that the series $\sum_{k = 1}^{\infty} (a_{k} + b_{k})$ diverges.

Leila finds that there are more factors affecting the number of salmon that return to Redfish Lake than the dams: There are good years and bad years. These happen at random, but they are more or less cyclical, so she models the number of fish $q_{k}$ returning each year as $q_{k + 1} = (0.14 {(- 1)}^{k} + 0.36) (q_{k} + h)$ , where h is the number of fish whose spawn she releases from the hatchery annually.

(a) Show that the sustained number of fish returning in even-numbered years approach approximately $q_{e} = 3 h \sum_{k = 1}^{\infty} 0 . 11^{k} .$

(Hint: Make a new recurrence by using two steps of the one given.)

(b) Show that the sustained number of fish returning in odd-numbered years approaches approximately $q_{o} = \frac{61}{11} h \sum_{k = 1}^{\infty} 0 . 11^{k} .$

(c) How should Leila choose h, the number of hatchery fish to breed in order to hold the minimum number of fish returning in each run near some constant P?

Express each of the repeating decimals in Exercises 71–78 as a geometric series and as the quotient of two integers reduced to lowest terms.

$0.6345345 . . .$

Use any convergence test from this section or the previous section to determine whether the series in Exercises 31–48 converge or diverge. Explain how the series meets the hypotheses of the test you select.

$\sum_{k = 1}^{\infty} \frac{1}{2 k + 7}$

Let $α$ be any real number. Show that there is a rearrangement of the terms of the alternating harmonic series that converges to $α$ . (Hint: Argue that if you add up some finite number of the terms of $\sum_{k = 1}^{\infty} \frac{1}{2 k - 1}$ , the sum will be greater than $α$ . Then argue that, by adding in some other finite number of the terms of

$\sum_{k = 1}^{\infty} (- \frac{1}{2 k})$ , you can get the sum to be less than $α$ . By alternately adding terms from these two divergent series as described in the preceding two steps, explain why the sequence of partial sums you are constructing will converge to $α$ .)

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Simplify the quotients in Exercises 21–28 without using a calculator.
$\frac{4!}{(2!)!}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of the expression.

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Most popular questions from this chapter

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Simplify the quotients in Exercises 21–28 without using a calculator.4!(2!)!

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of the expression.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Applied Mathematics

Calculus

Theoretical and Mathematical Physics

Geometry

Decision Maths

Study anywhere. Anytime. Across all devices.

Simplify the quotients in Exercises 21–28 without using a calculator.
$\frac{4!}{(2!)!}$