#$&* course MTH272 If your solution to a stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
.............................................
Given Solution: `a 1 / ( (x+1)(500-x) ) = A / (x+1) + B / (500-x) so A(500-x) + B(x+1) = 1 so A = 1 / 501 and B = 1/501. The integrand becomes 5010 [ 1 / (501 (x+1) ) + 1 / (501 (500 - x)) ] = 10 [ 1/(x+1) + 1 / (500 - x) ]. Thus we have t = INT(5010 ( 1 / (x+1) + 1 / (500 - x) ) , x). Since an antiderivative of 1/(x+1) is ln | x + 1 | and an antiderivative of 1 / (500 - x) is - ln | 500 - x | we obtain t = 10 [ ln (x+1) - ln (500-x) + c]. (for example INT (1 / (500 - x) dx) is found by first substituting u = 1 - x, giving du = -dx. The integral then becomes INT(-1/u du), giving us - ln | u | = - ln | 500 - x | ). We are given that x = 1 yields t = 0. Substituting x = 1 into the expression t = 10 [ ln (x+1) - ln (500-x) + c], and setting the result equal to 0, we get c = 5.52, appxox. So t = 10 [ ln (x+1) - ln (500-x) + 5.52] = 10 ln( (x+1) / (500 - x) ) + 55.2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ************************************************************ ********************************************* Question: `qHow long does it take for 75 percent of the population to become infected? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 75% of the 500 individuals susceptible to the disease = 375 [(10)ln(375+1))/(500-375))] + 55.2]=75% of population [(10)ln(375+1))/(500-375))] + 55.2]=66.2 Hours confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 75% of the population of 500 is 375. Setting x = 375 we get t = 10 ln( (375+1) / (500 - 375) ) + 55.2 = 66.2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* ********************************************* Question: `qWhat integral did you evaluate to obtain your result? The integral where 1 individual was equal to zero time [(10)ln((x+1)/(500-x))] + c]=t &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ****************************************************** ********************************************* Question: `qHow many people are infected after 100 hours? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [(10)ln((x+1)/(500-x))] + 55.2]=100 [(10)ln((x+1)/(500-x))]=44.8 subtract 55.2 from both sides ln((x+1)/(500-x))=4.48 divide both sides by 10 e^(ln((x+1)/(500-x)))=e^4.48 taking each side to the power of e (x+1)/(500-x)=88.2347 x+1=44117.35-88.2347x multiply each side by (500-x) and rearrange x+89.2347x=44117.35 divide by 89.2347 x=494.4 or 494 individuals will be infected in 100 hours since you can’t have .4 people confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a t = 10 ln( (x+1) / (500 - x) ) + 55.2. To find x we solve this equation for x: 10 ln( (x+1) / (500 - x) ) = t - 55.2 so ln( (x+1) / (500 - x) ) = (t - 55.2 ) / 10. Exponentiating both sides we have (x+1) / (500 - x) = e^( (t-55.2) / 10 ). Multiplying both sides by 500 - x we have x + 1 = e^( (t-55.2) / 10 ) ( 500 - x) so x + 1 = 500 e^( (t-55.2) / 10 ) - x e^( (t-55.2) / 10 ). Rearranging we have x + x e^( (t-55.2) / 10 ) = 500 e^( (t-55.2) / 10 ) - 1. Factoring the left-hand side x ( 1 + e^( (t-55.2) / 10 ) ) = 500 e^( (t-55.2) / 10 ) - 1 so that x = (500 e^( (t-55.2) / 10 ) - 1) / ( 1 + e^( (t-55.2) / 10 ) ). Plugging t = 100 into this expression we actually get x = 494.4, approx. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!