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guidelines and your promise that you will abide by them. GEOG1016 Nature Conservation for Sustainable Societies Professor C Y Jim Department of Geography The University of Hong Kong GEOG1016 Nature Conservation for Sustainable Societies / Professor C Y Jim 1 AN OVERVIEW OF NATURAL RESOURCES INTRODUCTION ENVIRONMENTAL CRISIS QUALITY OF LIFE NATURE OF RESOURCES CLASSIFICATION OF RESOURCES GEOG1016 Nature Conservation for Sustainable Societies / Professor C Y Jim 2 1 INTRODUCTION Human versus resources a resource humans give an object
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is the sum of 8 and y? | 8 + y | Express the number (x) of apples increased by two | x + 2 | Express the total weight of Alphie the dog (x) and Cyrus the cat (y) | x + y | Key words for Subtraction - less than‚ fewer than‚ reduced by‚ decreased by‚ difference of | What is four less than y | y - 4 | What is nine less than a number (y) | y - 9 | What if the number (x) of pizzas was reduced by 6? | x - 6 | What is the difference of my weight (x) and your weight (y) | x - y | Key words for multiplication *
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PROBABILITY P(A U B)= P(A) + P(B) – P(A∩B) If P(A∩B) = 0 then A and B are mutually exclusive and P(AUB) = P(A) + P(B) Joint Probability Marginal Probability PXY(x‚y) = P(X=x ∩ Y=y) PX(x) = ∑P(X=x ∩ Y=y) (For all values of y) Quotient Rule: Multiplication Rule P(A|B) = P(A∩B) / P(B) P(A∩B) = P(A|B) x P(B) = P(B|A) x P(A) Two events are statistically independent if: P(A|B) = P(A) P(B|A) = P(B) P(A∩B) = P(A) P(B) _ _ P(A) = P(A|B)P(B) + P(A|B)P(B) Bayes Rule:
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not defined 0 not defined 0 Domain and range of various trigonometric functions: Function Domain Range y = sin x π π − 2 ‚ 2 [–1‚ 1] y = cos x 0‚ π [–1‚ 1] π π − 2 ‚ 2 − {0} π 0‚ π − 2 y = cosec x y = sec x R – (–1‚ 1) π π − 2 ‚ 2 ( 0‚ π ) y = tan x y = cot x 10. R – (–1‚1) R R Sign Convention I II III IV sin x + + – – cos x + –
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4 4.1 Solving Systems of Linear Equations Graphically and Numerically 4.2 Solving Systems of Linear Equations by Substitution 4.3 Solving Systems of Linear Equations by Elimination 4.4 Systems of Linear Inequalities Systems of Linear Equations in Two Variables We can do anything we want to do if we stick to it long enough. —HELEN KELLER mericans have been moving toward a more mobile lifestyle. In recent years‚ the percentage of U.S. households relying solely on mobile phone service has
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CHENNAI INSTITUTE OF TECHNOLOGY Sarathy Nagar‚ Kundrathur‚ Pudupedu‚ Chennai– 600 069. Department of Mathematics Subject Name: Numerical Methods Subject Code: MA1251 Unit I 1) Write the Descartes rule of signs Sol: 1) An equation f ( x) = 0 cannot have more number of positive roots than there are changes of sign in the terms of the polynomial f ( x) . 2)An equation f ( x) = 0 cannot have more number of positive roots than there are changes of sign in the terms of the polynomial f
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example‚ y = 5y and xy − sin x = 0 are first order linear d.e.; (y )2 + (y )5 − y = ex is third order‚ nonlinear. We observe that in general‚ a d.e. has many solutions‚ e.g. y = sin x + c‚ c an arbitrary constant‚ is a solution of y = cos x. Such solutions containing arbitrary constants are called general solution of a given d.e.. Any solution obtained from the general solution by giving specific values to the arbitrary constants is called a particular solution of that d.e. e.g. 2 y = sin x +
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Rock it out loud: discovering the location of harmonies that make your guitar sing. Purpose: To identify the locations of harmonics on an acoustic guitar and relate them to guitar string lengths. Hypothesis: If I pluck the sixth string of the guitar from frets nine- twelve‚ then those frets will not produce harmonics that are able to be heard clearly. Materials: Acoustic guitar ( adult or child sized) Cloth tape measure‚ metric Lab Notebook Procedures on finding harmonics on your
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c). (a) y = y 2 (b) y = 1 + y 2 (c) yy = 3x (d) y = tan(x + C) (e) 3x2 − y 2 = C (f) y = −1/(x + C) 2. Find the value of k so that y = e3t + ke2t is a solution of y − 2y − 3y = 3e2t . 3. Solve the following differential equations and IVP’s. You may solve these equations implicitly. (a) y + 3x2 y = x2 (b) y ln t = y+1 t 2 dy dt 2 sin t dy = 0 y (d) y y − t = 0‚ y(1) = 2‚ y (1) = 1 (c) cos t dt + 1 + (e) y − y = et y 2 (f) cos(xy) − xy sin(xy) + 2xyex + (ex − 2y − x2 sin(xy))y = 0‚ y(1) = 0
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