II. Given:Find:Pr (D)=.25Pr (G and D)Pr(G) = .2Pr(D/G)Pr (G/D) = .05Pr (G and L)Pr(L) =, 15Pr (D and L)

Answered: Image /qna-images/answer/b2614461-3ad0-41a9-909a-be850f1ad3f2.jpg

Answered: II. Given: Find: Pr (D) = .25 Pr (G and…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Pierce Manufacturing determines that the daily revenue, in dollars, from the sale of x lawn chairs is R(x) = 0.003x³ +0.03x² +0.4x. Currently, Pierce sells 70 lawn chairs daily. a) What is the current daily revenue? b) How much would revenue increase if 74 lawn chairs were sold each day? c) What is the marginal revenue when 70 lawn chairs are sold daily? d) Use the answer from part (c) to estimate R(71), R(72), and R(73). C a) The current revenue is $ b) The revenue would increase by $ (Round to the nearest cent.) c) The marginal revenue is $ when 70 lawn chairs are sold daily. d) R(71) $ R(72) $ R(73) $
(a) Construct a scatterplot of the diastolic blood pressure (y) versus the systolic blood pressure (x).
Use a linear approximation to estimate (1.03) ³ (-1.01) ² Compute the percentage error: %
P(A)= 0.23, P(B)= 0.46, P(A and B)= 0.15 Calculate P(A' and B)
Pierce Manufacturing determines that the daily revenue, in dollars, from the sale of x lawn chairs is R(x)=0.005x +0.04x²+0.4x Currently, Pierce sells 50 lawn chairs daily. a) What is the current daily revenue? b) How much would revenue increase if 54 lawn chairs were sold each day? c) What is the marginal revenue when 50 lawn chairs are sold daily? d) Use the answer from part (c) to estimate R(51), R(52), and R(53). a) The current revenue is $ b) The revenue would increase by $ c) The marginal revenue is $when d) R(51) $ S R(52) S R(53) $ (Round to the nearest cent.) 50 lawn chairs are sold daily
P(A)= 0.33, P(B)= 0.44, P(A and B)= 0.16 Calculate P(A and B')?
(i) Using the integral definition below, find the average density of the Earth given the following information: Oceanic Plate where can be written as 3-5 miles thick ● Diagram not drawn to scale 1400 miles Thick 1800 miles thick 5-25 miles thick 800 miles thick Continental Plate Mantle (Flows with the consistancy of asphalt) Scale everything in terms of the radius of the Earth R ( 4000 miles here for simplicity), so that in terms of the variable y r/R, 0 ≤ y ≤ 1, the integral for the average density p=2³3r²p(r)dr Outer Core (Iron and nickel in the liquid state) Inner Core (Iron and nickel in the solid state} p=3√ 1²p(v)dy, p(y) = P₁, 0≤ y ≤yi, = P2, y1 ≤y ≤y2, = P3, y2 ≤y ≤ y3, = P4, Y3 ≤ y ≤ 1. INNER CORE (0 ≤ y ≤ yı): Earth's inner core has the highest density p₁ 13 g/cm³. OUTER CORE (1 ≤ y ≤ y2): Next, the outer core has a density P2 11 g/cm³. LOWER MANTLE (y2 ≤ y ≤y3): The lower mantle has a significantly lower density P35 g/cm³. This is about 1400 miles thick.
Between Within Source Total SS 48.56 115.62 df 15 17 MS F eta² ||
The number of red lights, Y, that Sam stops at on his way to work is a random variable with probability distribution given by: (a) Find the value of a. Y 0 1 විය. 2 3 P(Y= y) 3a 8a² 4a² 8a
P(A)=0.71 P(B)こo.5 z P(BUA)=D0.85 3u P(AnB)= P(BIA)= PLAIB)= %3D

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II. Given: Find: Pr (D) = .25 Pr (G and D) Pr(G)=.2 Pr(D/G) Pr (G/D) = .05 Pr (G and L) Pr(L)=, 15 Pr (D and L)