University:
Amherst College
Course:
STAT 111 | Introduction to Statistics
Academic year:

2022
Views:

328

Pages:

7
Author:

ulbistasdeba3xr8

Answers Sex. 1 the # of people is 2.93 x 10^8 of square miles is 3.54 x 10^6 so the density is n = 2.93 x 10^8 / 3.54 x 10^6 = 82.8 people / square mile the web says dist from Denver to Cheyenne = 98 miles to Oklahoma City = 509 miles and the # of people living in the two cities is Cheyenne 2.5 million Okl. City 67.4 million the definition of a light year is the distance light travels in a year - so the distance light travels in a million years is a million light years so it's correcty to think of 2 ways to find Method 1: Simplest calculations. Take the data points, say the 1st & last. First: N = 4.9 x 10^9 = 9 + 1000^6 mat (1000)^3 Next: N = 1400 x 10^9 = 9 + 4000^6 mat (4000)^3 i.e. 4.9 = 9 + 1000^6 7.5 = 9 + 4000^6 during those 2 years ---> 4.03 x 10^6 10 = 8.67 x 10^24 dist D N N/D^3 1000 4.90E+09 4.9 2000 4.60E+10 5.75 3000 1.60E+11 5.92592593 4000 4.80E+11 7.5 8 7 6 5 4 3 2 1 0 0 1000 2000 3000 4000 5000 excel fits = trend line with y = 0.0008x + 4.026 b = 4.026 = 8 pi / c^3 these agree pretty well with the first method. But notice that they do not agree exactly. If we wanted very accurate answers we would have to think about why, and which method is the most accurate to conclusion we know: C = 4πr / 3 => n = 3C / 4πr so b = πr² = πn => n = b / πr I will use: n = 0.96 Fermi/s / (million light years)^3 = 2.55 x 10^-4 Fermi/s / (million light years)^3 * (million years)^3 (d) t = n/n = 0.96 million years / 2.55 x 10^-4 t = 3760 million years = 3.76 billion years Cepheids we know F = L/4πR² = c/4πR² L = 4πR²F = 4πR²c/400d R = 57 light years R = 5.37 x 10^19 cm so L #1 6.14 x 10^35 erg/sec #2 9.87 x 10^35 erg/sec #3 1.83 x 10^36 erg/sec 160 258 477 the 7-day cepheid most here a Luminosity ~ 385 Lsun, or 1.47 x 10^36 erg/sec. It's distance is R = [L/4πF]^(1/2) = 4.25 x 10^24 cm = 4.47 million light years 6.5 fl -15 erg/cm^2 sec

Answer Set 1

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