BIG IDEAS MATH Integrated Math 1: Student Edition 2016
16th Edition
ISBN: 9781680331127
Author: HOUGHTON MIFFLIN HARCOURT
Publisher: Houghton Mifflin Harcourt
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Question
Chapter 4.5, Problem 29E
(a)
To determine
To predict: The total number of people reported
(a)
Expert Solution
Answer to Problem 29E
The total number of people reported an earthquake
The total number of people reported an earthquake
Explanation of Solution
Given:
| Minutes, x | ||
| People, y |
Calculation:
Substituting
Therefore, using the model
The total number of people reported an earthquake
The total number of people reported an earthquake
Conclusion:
Therefore, the total number of people reported an earthquake
The total number of people reported an earthquake
(b)
To determine
To describe: The accuracy of the extrapolations
(b)
Expert Solution
Answer to Problem 29E
The second prediction was less accurate than the first
Explanation of Solution
Given:
| Minutes, x | ||
| People, y |
Calculation:
The difference between the first extrapolation and the actual data value is only
So, the second prediction was less accurate than the first. This fits what you would expect because the farter removed a value if from the known values, the less confidence you can have in the accuracy of prediction
Conclusion:
The second prediction was less accurate than the first
Chapter 4 Solutions
BIG IDEAS MATH Integrated Math 1: Student Edition 2016
Ch. 4.1 - Prob. 1ECh. 4.1 - Prob. 2ECh. 4.1 - Prob. 3ECh. 4.1 - Prob. 4ECh. 4.1 - Prob. 5ECh. 4.1 - Prob. 6ECh. 4.1 - Prob. 7ECh. 4.1 - Prob. 8ECh. 4.1 - Prob. 9ECh. 4.1 - Prob. 10E
Ch. 4.1 - Prob. 11ECh. 4.1 - Prob. 12ECh. 4.1 - Prob. 13ECh. 4.1 - Prob. 14ECh. 4.1 - Prob. 15ECh. 4.1 - Prob. 16ECh. 4.1 - Prob. 17ECh. 4.1 - Prob. 18ECh. 4.1 - Prob. 19ECh. 4.1 - Prob. 20ECh. 4.1 - Prob. 21ECh. 4.1 - Prob. 22ECh. 4.1 - Prob. 23ECh. 4.1 - Prob. 24ECh. 4.1 - Prob. 25ECh. 4.1 - Prob. 26ECh. 4.1 - Prob. 27ECh. 4.1 - Prob. 28ECh. 4.1 - Prob. 29ECh. 4.1 - Prob. 30ECh. 4.1 - Prob. 31ECh. 4.1 - Prob. 32ECh. 4.1 - Prob. 33ECh. 4.1 - Prob. 34ECh. 4.1 - Prob. 35ECh. 4.1 - Prob. 36ECh. 4.1 - Prob. 37ECh. 4.1 - Prob. 38ECh. 4.1 - Prob. 39ECh. 4.1 - Prob. 40ECh. 4.1 - Prob. 41ECh. 4.1 - Prob. 42ECh. 4.1 - Prob. 43ECh. 4.1 - Prob. 44ECh. 4.1 - Prob. 45ECh. 4.2 - Prob. 1ECh. 4.2 - Prob. 2ECh. 4.2 - Prob. 3ECh. 4.2 - Prob. 4ECh. 4.2 - Prob. 5ECh. 4.2 - Prob. 6ECh. 4.2 - Prob. 7ECh. 4.2 - Prob. 8ECh. 4.2 - Prob. 9ECh. 4.2 - Prob. 10ECh. 4.2 - Prob. 11ECh. 4.2 - Prob. 12ECh. 4.2 - Prob. 13ECh. 4.2 - Prob. 14ECh. 4.2 - Prob. 15ECh. 4.2 - Prob. 16ECh. 4.2 - Prob. 17ECh. 4.2 - Prob. 18ECh. 4.2 - Prob. 19ECh. 4.2 - Prob. 20ECh. 4.2 - Prob. 21ECh. 4.2 - Prob. 22ECh. 4.2 - Prob. 23ECh. 4.2 - Prob. 24ECh. 4.2 - Prob. 25ECh. 4.2 - Prob. 26ECh. 4.2 - Prob. 27ECh. 4.2 - Prob. 28ECh. 4.2 - Prob. 29ECh. 4.2 - Prob. 30ECh. 4.2 - Prob. 31ECh. 4.2 - Prob. 32ECh. 4.2 - Prob. 33ECh. 4.2 - Prob. 34ECh. 4.2 - Prob. 35ECh. 4.2 - Prob. 36ECh. 4.2 - Prob. 37ECh. 4.2 - Prob. 38ECh. 4.2 - Prob. 39ECh. 4.2 - Prob. 40ECh. 4.2 - Prob. 41ECh. 4.2 - Prob. 42ECh. 4.2 - Prob. 43ECh. 4.2 - Prob. 44ECh. 4.3 - Prob. 1ECh. 4.3 - Prob. 2ECh. 4.3 - Prob. 3ECh. 4.3 - Prob. 4ECh. 4.3 - Prob. 5ECh. 4.3 - Prob. 6ECh. 4.3 - Prob. 7ECh. 4.3 - Prob. 8ECh. 4.3 - Prob. 9ECh. 4.3 - Prob. 10ECh. 4.3 - Prob. 11ECh. 4.3 - Prob. 12ECh. 4.3 - Prob. 13ECh. 4.3 - Prob. 14ECh. 4.3 - Prob. 15ECh. 4.3 - Prob. 16ECh. 4.3 - Prob. 17ECh. 4.3 - Prob. 18ECh. 4.3 - Prob. 19ECh. 4.3 - Prob. 20ECh. 4.3 - Prob. 21ECh. 4.3 - Prob. 22ECh. 4.3 - Prob. 23ECh. 4.3 - Prob. 24ECh. 4.3 - Prob. 25ECh. 4.3 - Prob. 26ECh. 4.3 - Prob. 27ECh. 4.3 - Prob. 28ECh. 4.3 - Prob. 29ECh. 4.3 - Prob. 30ECh. 4.3 - Prob. 31ECh. 4.3 - Prob. 32ECh. 4.3 - Prob. 33ECh. 4.3 - Prob. 34ECh. 4.3 - Prob. 35ECh. 4.3 - Prob. 36ECh. 4.3 - Prob. 37ECh. 4.3 - Prob. 38ECh. 4.3 - Prob. 1QCh. 4.3 - Prob. 2QCh. 4.3 - Prob. 3QCh. 4.3 - Prob. 4QCh. 4.3 - Prob. 5QCh. 4.3 - Prob. 6QCh. 4.3 - Prob. 7QCh. 4.3 - Prob. 8QCh. 4.3 - Prob. 9QCh. 4.3 - Prob. 10QCh. 4.3 - Prob. 11QCh. 4.3 - Prob. 12QCh. 4.3 - Prob. 13QCh. 4.3 - Prob. 14QCh. 4.3 - Prob. 15QCh. 4.3 - Prob. 16QCh. 4.4 - Prob. 1ECh. 4.4 - Prob. 2ECh. 4.4 - Prob. 3ECh. 4.4 - Prob. 4ECh. 4.4 - Prob. 5ECh. 4.4 - Prob. 6ECh. 4.4 - Prob. 7ECh. 4.4 - Prob. 8ECh. 4.4 - Prob. 9ECh. 4.4 - Prob. 10ECh. 4.4 - Prob. 11ECh. 4.4 - Prob. 12ECh. 4.4 - Prob. 13ECh. 4.4 - Prob. 14ECh. 4.4 - Prob. 15ECh. 4.4 - Prob. 16ECh. 4.4 - Prob. 17ECh. 4.4 - Prob. 18ECh. 4.4 - Prob. 19ECh. 4.4 - Prob. 20ECh. 4.4 - Prob. 21ECh. 4.4 - Prob. 22ECh. 4.4 - Prob. 23ECh. 4.4 - Prob. 24ECh. 4.4 - Prob. 25ECh. 4.4 - Prob. 26ECh. 4.4 - Prob. 27ECh. 4.4 - Prob. 28ECh. 4.5 - Prob. 1ECh. 4.5 - Prob. 2ECh. 4.5 - Prob. 3ECh. 4.5 - Prob. 4ECh. 4.5 - Prob. 5ECh. 4.5 - Prob. 6ECh. 4.5 - Prob. 7ECh. 4.5 - Prob. 8ECh. 4.5 - Prob. 9ECh. 4.5 - Prob. 10ECh. 4.5 - Prob. 11ECh. 4.5 - Prob. 12ECh. 4.5 - Prob. 13ECh. 4.5 - Prob. 14ECh. 4.5 - Prob. 15ECh. 4.5 - Prob. 16ECh. 4.5 - Prob. 17ECh. 4.5 - Prob. 18ECh. 4.5 - Prob. 19ECh. 4.5 - Prob. 20ECh. 4.5 - Prob. 21ECh. 4.5 - Prob. 22ECh. 4.5 - Prob. 23ECh. 4.5 - Prob. 24ECh. 4.5 - Prob. 25ECh. 4.5 - Prob. 26ECh. 4.5 - Prob. 27ECh. 4.5 - Prob. 28ECh. 4.5 - Prob. 29ECh. 4.5 - Prob. 30ECh. 4.5 - Prob. 31ECh. 4.5 - Prob. 32ECh. 4.5 - Prob. 33ECh. 4.6 - Prob. 1ECh. 4.6 - Prob. 2ECh. 4.6 - Prob. 3ECh. 4.6 - Prob. 4ECh. 4.6 - Prob. 5ECh. 4.6 - Prob. 6ECh. 4.6 - Prob. 7ECh. 4.6 - Prob. 8ECh. 4.6 - Prob. 9ECh. 4.6 - Prob. 10ECh. 4.6 - Prob. 11ECh. 4.6 - Prob. 12ECh. 4.6 - Prob. 13ECh. 4.6 - Prob. 14ECh. 4.6 - Prob. 15ECh. 4.6 - Prob. 16ECh. 4.6 - Prob. 17ECh. 4.6 - Prob. 18ECh. 4.6 - Prob. 19ECh. 4.6 - Prob. 20ECh. 4.6 - Prob. 21ECh. 4.6 - Prob. 22ECh. 4.6 - Prob. 23ECh. 4.6 - Prob. 24ECh. 4.6 - Prob. 25ECh. 4.6 - Prob. 26ECh. 4.6 - Prob. 27ECh. 4.6 - Prob. 28ECh. 4.6 - Prob. 29ECh. 4.6 - Prob. 30ECh. 4.6 - Prob. 31ECh. 4.6 - Prob. 32ECh. 4.6 - Prob. 33ECh. 4.6 - Prob. 34ECh. 4.6 - Prob. 35ECh. 4.6 - Prob. 36ECh. 4.6 - Prob. 37ECh. 4.6 - Prob. 38ECh. 4.6 - Prob. 39ECh. 4.6 - Prob. 40ECh. 4.6 - Prob. 41ECh. 4.6 - Prob. 42ECh. 4.6 - Prob. 43ECh. 4.6 - Prob. 44ECh. 4.6 - Prob. 45ECh. 4.6 - Prob. 46ECh. 4.6 - Prob. 47ECh. 4.6 - Prob. 48ECh. 4.6 - Prob. 49ECh. 4.6 - Prob. 50ECh. 4.6 - Prob. 51ECh. 4.6 - Prob. 52ECh. 4.6 - Prob. 53ECh. 4.6 - Prob. 54ECh. 4.6 - Prob. 55ECh. 4.6 - Prob. 56ECh. 4.6 - Prob. 57ECh. 4.6 - Prob. 58ECh. 4.6 - Prob. 59ECh. 4.6 - Prob. 60ECh. 4.6 - Prob. 61ECh. 4.6 - Prob. 62ECh. 4.6 - Prob. 63ECh. 4.6 - Prob. 64ECh. 4.6 - Prob. 65ECh. 4 - Prob. 1CRCh. 4 - Prob. 2CRCh. 4 - Prob. 3CRCh. 4 - Prob. 4CRCh. 4 - Prob. 5CRCh. 4 - Prob. 6CRCh. 4 - Prob. 7CRCh. 4 - Prob. 8CRCh. 4 - Prob. 9CRCh. 4 - Prob. 10CRCh. 4 - Prob. 11CRCh. 4 - Prob. 12CRCh. 4 - Prob. 13CRCh. 4 - Prob. 14CRCh. 4 - Prob. 15CRCh. 4 - Prob. 16CRCh. 4 - Prob. 17CRCh. 4 - Prob. 18CRCh. 4 - Prob. 19CRCh. 4 - Prob. 20CRCh. 4 - Prob. 21CRCh. 4 - Prob. 22CRCh. 4 - Prob. 23CRCh. 4 - Prob. 24CRCh. 4 - Prob. 25CRCh. 4 - Prob. 26CRCh. 4 - Prob. 27CRCh. 4 - Prob. 28CRCh. 4 - Prob. 29CRCh. 4 - Prob. 30CRCh. 4 - Prob. 1CTCh. 4 - Prob. 2CTCh. 4 - Prob. 3CTCh. 4 - Prob. 4CTCh. 4 - Prob. 5CTCh. 4 - Prob. 6CTCh. 4 - Prob. 7CTCh. 4 - Prob. 8CTCh. 4 - Prob. 9CTCh. 4 - Prob. 10CTCh. 4 - Prob. 11CTCh. 4 - Prob. 12CTCh. 4 - Prob. 13CTCh. 4 - Prob. 1CACh. 4 - Prob. 2CACh. 4 - Prob. 3CACh. 4 - Prob. 4CACh. 4 - Prob. 5CACh. 4 - Prob. 6CACh. 4 - Prob. 7CACh. 4 - Prob. 8CACh. 4 - Prob. 9CA
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- By considering appropriate series expansions, e². e²²/2. e²³/3. .... = = 1 + x + x² + · ... when |x| < 1. By expanding each individual exponential term on the left-hand side the coefficient of x- 19 has the form and multiplying out, 1/19!1/19+r/s, where 19 does not divide s. Deduce that 18! 1 (mod 19).arrow_forwardProof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.arrow_forwardBy considering appropriate series expansions, ex · ex²/2 . ¸²³/³ . . .. = = 1 + x + x² +…… when |x| < 1. By expanding each individual exponential term on the left-hand side and multiplying out, show that the coefficient of x 19 has the form 1/19!+1/19+r/s, where 19 does not divide s.arrow_forward
- Let 1 1 r 1+ + + 2 3 + = 823 823s Without calculating the left-hand side, prove that r = s (mod 823³).arrow_forwardFor each real-valued nonprincipal character X mod 16, verify that L(1,x) 0.arrow_forward*Construct a table of values for all the nonprincipal Dirichlet characters mod 16. Verify from your table that Σ x(3)=0 and Χ mod 16 Σ χ(11) = 0. x mod 16arrow_forward
- For each real-valued nonprincipal character x mod 16, verify that A(225) > 1. (Recall that A(n) = Σx(d).) d\narrow_forward24. Prove the following multiplicative property of the gcd: a k b h (ah, bk) = (a, b)(h, k)| \(a, b)' (h, k) \(a, b)' (h, k) In particular this shows that (ah, bk) = (a, k)(b, h) whenever (a, b) = (h, k) = 1.arrow_forward20. Let d = (826, 1890). Use the Euclidean algorithm to compute d, then express d as a linear combination of 826 and 1890.arrow_forward
- Let 1 1+ + + + 2 3 1 r 823 823s Without calculating the left-hand side, Find one solution of the polynomial congruence 3x²+2x+100 = 0 (mod 343). Ts (mod 8233).arrow_forwardBy considering appropriate series expansions, prove that ez · e²²/2 . e²³/3 . ... = 1 + x + x² + · ·. when <1.arrow_forwardProve that Σ prime p≤x p=3 (mod 10) 1 Р = for some constant A. log log x + A+O 1 log x ,arrow_forward
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