And length of minor axis = 2b = 2*10 = 20. c) From part a) and b), we have a = 8 and b = 10 and h=3,k=-5 So, c2=b2-a2=102-82=100-64=36 =>c=sqrt36= 6. In this given equation b>a So, this is a vertical ellipse. Therefore, coordinates of foci = (3, -5+6) and (3, -5-6) Foci are (3,1) and (3, -11). d) This is the graph of given ellipse with foci and center. (12 points) Score | | 2.
Take that difference and divide by the actual value (given in this problem) of tin’s atomic radius. Take your answer and multiply by 100 to make it a percentage, and that is your percent error. Percent error is a common calculation in science. Your grade will not be determined by the amount of error, but just that you used the calculation correctly. (140-145)/ 140 x = 3.5 6.
The graph derivative vs. Diameter was linear because r^2 value was closer to 1. 7. The units of the derivative vs. diameter graph could represent _______________ 8. Mass vs.
Name: ______________________ Class: _________________ Date: _________ ID: A quiz 6.1-6.3 Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. ____ ____ 1. Determine which binomial is a factor of −2x 3 + 14x 2 − 24x + 20. a. x + 5 b. x + 20 c. x – 24 d. x – 5 2. The volume of a shipping box in cubic feet can be expressed as the polynomial 2x 3 + 11x 2 + 17x + 6. Each dimension of the box can be expressed as a linear expression with integer coefficients.
TOPIC 8 Chi-Square goodness-of-fit test Problem 12.1 Use a chi-square goodness-of-fit to determine whether the observed frequencies are distributed the same as the expected frequencies (α = .05) Category | fo | fe | 1 | 53 | 68 | 2 | 37 | 42 | 3 | 32 | 33 | 4 | 28 | 22 | 5 | 18 | 10 | 6 | 15 | 8 | Step 1 Ho: The observed frequencies are distributed the same as the expected frequencies Ha: The observed frequencies are not distributed the same as the expected frequencies Step 2 df = k – m – 1 Step 3 α = 0.05 x 2 0.05, 5df = 11.0705 Step 4 Reject Ho if x 2 > 11.0705 Category | fo | fe | | 1 | 53 | 68 | | 2 | 37 | 42 | | 3 | 32 | 33 | | 4 | 28 | 22 | | 5 | 18 | 10 | | 6 | 15 | 8
The distance between the forces is given by the Coulombs law through the use of the formula F=kq1q2/r2.0.1newtons = 8.99*109*3.2*10-6*7.7*10-7/r2 R= 555.78 Answer to question 3 • Potential difference between the two plates is equal to velocity which is equal to 6.0*106m/s • Force = mass *acceleration = 1.4*10-13*6.0*106 = -8254 nektons The speed of the particles are computed by the formula V=ED. This is equal to 8.5*10-6*0.15. This is equal to 84.1 Answer to question 4 Voltage = current *resistance. This implies that in this case while V is 5.0 and resistance is 1.0*103, current will be equal to 5/1.0*103, = 500 amps B the direction of the conventional current provides the electric charge movement from the positive side of the battery to its negative side as in indicated in the diagram below Answer to question 5 • This section focuses on the equivalent resistance of a circuit. The equivalent resistance will be equal to (5.0*102+1.00*103)2.
160.131 Mathematics for Business (1) Test 1 4th of April, 2012 Formulas you may need Slopes (Gradients) The slope (or gradient) m of a curve measures the rate of change in the dependent variable (y), i.e. m = [pic] = [pic] = [pic] = [pic] Determining the Equation of a Line: If we know 2 distinct points (x1, y1) and (x2, y2) on the line: y – y1 = [pic](x – x1); if we know a point (x0, y0) and the slope m of the line: y – y0 = m(x – x0); if we know the slope m and the y-intercept (0, c) of the line: y = mx + c. multiplier = - [pic] Inverse Matrices for 2 × 2 matrices: If A = [pic] then A-1 = [pic][pic], provided ad – bc [pic] 0. Best Fit Curves and
In what sense is the Newtonian universe simpler than Ptolemy’s? Suppose observations had shown that the two did equally well at explaining the data. Construct an argument to say that Newton’s universe should still be preferred. The idea that the universe works mechanically, like a clock. It is called Newtonian, because it began from the Newtonian discovery of the laws of gravity and world motion.
2) The larger moveable knife edge was then clamped to the pendulum, at a small distance (1cm) above the centre of mass. 3) The distance h was then determined from the centre of mass to the axis of suspension. 4) The mask was then attached to the pendulum. 5) The light gate was then connected to the digit-metre and adjust its height and position relative to the mask to allow the period T, of the pendulum to be measured repeatedly. 6) The procedure was then repeated for larger values of h, until T has passed its minimum value.
Through both qualitative observation, and our quantitative data, we deduced that the shorter the length of a pendulum, the smaller its period. At the longest length 42cm, the period was 1.33s. Each progressing length: 33.5cm, 30cm, 22cm, and 11cm yielded smaller and smaller periods of 1.16s, 1.15s, 0.96s, and 0.68s respectively. In our first graph of Period vs. Length, we plotted the points and discovered that it formed a root curve.