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# Recent problems solved by 'stanbon'

Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 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 percentage/457205: Interest question, please help \$5400 is invested, part of it at 10% and part of it at 7%. For a certain year the total yield is \$468. How much was invested at each rate? Thanks for your help.1 solutions Answer 313703 by stanbon(57347)   on 2011-06-03 20:08:27 (Show Source): You can put this solution on YOUR website!Interest question, please help \$5400 is invested, part of it at 10% and part of it at 7%. For a certain year the total yield is \$468. How much was invested at each rate? ----------------- Quantity Equation: x + y = 5400 dollars Interest Equation:0.10x + 0.07y = 469 dollars ------------------------------------------- Multiply thru 1st by 10 Multiply thru 2nd by 100 -------------------------------- 10x + 10y = 54000 10x + 7y = 46900 --------------------------- Subtract and solve for "y": 3y = 7100 y = \$2366.67 (amt. invested at 7%) --- Solve for "x": x+ y = 5400 x = 5400-2366.67 x = \$2633.33 (amt. invested at 10%) ======================================= Cheers, Stan H. =============
 Coordinate-system/457208: I can't seem to get this Write an equation in standard form for a line passing through the pair of points. ( 7, 8) and ( 7, -8) Any help would be appreciated1 solutions Answer 313702 by stanbon(57347)   on 2011-06-03 20:02:25 (Show Source): You can put this solution on YOUR website!( 7, 8) and ( 7, -8) ----------------- slope = (-8-8)/(7-7) = -16/0 which is "not defined". ---- Notice that both points have x=7. Plot those two points and you will get a vertical line passing thru all the points where x=7 -------------------- Equation of the line: x = 7 =============================== Cheers, Stan H. ==========
 Complex_Numbers/457212: How can I simplify 1+2i/2-3i1 solutions Answer 313701 by stanbon(57347)   on 2011-06-03 19:58:49 (Show Source): You can put this solution on YOUR website!simplify 1+2i/2-3i ---- Multiply numerator and denominator by the conjugate of 2-3i, which is 2+3i. ---- You get: [(1+2i)(2+3i)]/[(2-3i)(2+3i)] ==== = [2+5i-6]/[4 + 0i + 9] ------ = (-4+5i)/13 ================ Cheers, Stan H. ================
 Polynomials-and-rational-expressions/457210: Hi I am really confused on how to complete this problem. I am in a study plan and very stuck in a section. I need to know where they are getting certain numbers so i can complete the rest of these assignments... 9th rt b^2 * sqrt b^9 I get this b^2/9 b^9/2=b^2/18 b^9/18 from here I am confused. It shows in help that it is b^85/18. Where are they getting 85. The next confusion is after this it shows b^4+13/18. Where are they getting 4+13.. Can anyone help. It would be wonderful. Thank You1 solutions Answer 313699 by stanbon(57347)   on 2011-06-03 19:55:29 (Show Source): You can put this solution on YOUR website!9th rt b^2 * sqrt b^9 I get this b^2/9 b^9/2=b^2/18 b^9/18 ------------------- If what you say is correct you should add the exponents to get b^(11/9) --- from here I am confused. It shows in help that it is b^85/18. --- To get 85/18 you would nee 9/2 + 2/9 = (81+4)/18 = 85/18 --- Maybe your problem is not what you posted but rather: sqrt(b^9)* 9th rt. of b^2 which would give you b^(9/2) * b^(2/9) = b^(9/2 + 2/9) = b^(85/18) ============================================== Cheers, Stan H.
 Mixture_Word_Problems/457199: Another problem I need help with Suppose for a certain company C(x)= 35x + 300,000 represents the total function, and R(x)=55x represents the total revenue function. Find the total function and break even point.1 solutions Answer 313692 by stanbon(57347)   on 2011-06-03 18:56:03 (Show Source): You can put this solution on YOUR website!Suppose for a certain company C(x)= 35x + 300,000 represents the total cost function, and R(x)=55x represents the total revenue function. Find the total profit function and break even point. --------------------------------------- Profit = Revenue - Cost ---- P(x) = 55x -(35x+300,000) ----- P(x) = 20x - 300,000 ----------------------------- Break Even occurs when there is zero profit. ---- Solve 20x-300,000 = 0 20x = 300,000 x = 15000 ------------- Cheers, Stan H. ===============
 Quadratic_Equations/457191: How do you change a problem such as, 2x^2=-6+8x, into a quadratic equation that can be solved using the quadratic formula?1 solutions Answer 313690 by stanbon(57347)   on 2011-06-03 18:41:47 (Show Source): You can put this solution on YOUR website!2x^2=-6+8x, ------------------ Rearrange: 2x^2-8x+6 = 0 ---- Divide thru by 2 to get: x^2 - 4x + 3 = 0 ------------------------ Use the quadratic formula: x = [4 +- sqrt(16-4*3)]/2 ---- x = [4 +- sqrt(4)]/2 ---- x = [4 +- 2]/2 ---- x = (6/2) or x = (2/2) --- x = 3 or x = 1 ========================= Cheers, Stan H. ==================
 Linear-systems/457193: how do u solve this equation ok so here is the sentance complete the solution of t he eqation find the value of y when x=28 x+3y=28 please help me solve this problem!1 solutions Answer 313684 by stanbon(57347)   on 2011-06-03 18:26:25 (Show Source): You can put this solution on YOUR website!find the value of y when x=28 x+3y=28 please help me solve this problem! --------------- Substitute for "x" and solve for "y": ----- x+3y = 28 28 + 3y = 28 3y = 0 y = 0 ============ Cheers, Stan H.
 Trigonometry-basics/457189: use the law of sines to side b= 40cm angle A=120 angle C=58 1 solutions Answer 313681 by stanbon(57347)   on 2011-06-03 18:22:24 (Show Source): You can put this solution on YOUR website!use the law of sines to side b= 40cm angle A=120 angle C=58 -------------------- angle B = 180-(58+120) = 2 degrees --------------------------------------- Solve a/sin(120) = 40/sin(2) ---- Solve c/sin(58) = 40/sin(2) =============================== Cheers, Stan H.
 test/457190: There are a total 108 offensive players on a football team. There are twice as many runners than passing players. Write two equations and then slove for both equation. How many runners are there and how many passing people are there?1 solutions Answer 313680 by stanbon(57347)   on 2011-06-03 18:14:05 (Show Source): You can put this solution on YOUR website!There are a total 108 offensive players on a football team. There are twice as many runners than passing players. Write two equations and then solve for both equation. How many runners are there and how many passing people are there? --- Equations: r + p = 108 players r = 2p ---- Substitute for "r" and solve for "P": 2p + p = 108 3p = 108 p = 36 (# of passers) ---- solve for "r": r + p = 108 r + 36 = 108 r = 72 (# of runners) ======================== Cheers, Stan H.
 Probability-and-statistics/457084: The size of the box in a boxplot shows the ____________________ of the data set. A) difference between the mean and the median B) variance C) skewness D) interquartile range 1 solutions Answer 313660 by stanbon(57347)   on 2011-06-03 16:18:19 (Show Source): You can put this solution on YOUR website!The size of the box in a boxplot shows the ____ of the data set. ----------- Comment: The box shows the spread of data between Q1 and Q3. I don't like any of the option answers you have listed. ------------------------- Cheers, Stan H. ================ A) difference between the mean and the median ---------------- B) variance ---------------- C) skewness ---------------- D) interquartile range -------------------
 Probability-and-statistics/457143: The recovery rate from a flu is 0.9. If 4 people have this flu, what is the probability that: a) all will recover? b) at least 2 will recover? c) exactly 2 will recover? Assume independence. Thanks so much for your help!1 solutions Answer 313659 by stanbon(57347)   on 2011-06-03 16:12:36 (Show Source): You can put this solution on YOUR website!The recovery rate from a flu is 0.9. If 4 people have this flu, what is the probability that: a) all will recover? Ans: (0.9)^4 = -------------------------- b) at least 2 will recover? Ans: P(2<= x <=4) = 1 - P(0<= x <=2) = 1-binomcdf(4,0.9,2) = 1-0.0523 = 0.9477 ==================== c) exactly 2 will recover? Ans: P(x = 2) = 4C2(0.9)^2(0.1)^2 = binompdf(4,0.9,2) = 0.048600 ==================================================================== Cheers, Stan H. ================
 Probability-and-statistics/457075: Given the random variable X = {5, 6} with P(5) = 0.2 and P(6) = 0.8. Find E(X).1 solutions Answer 313658 by stanbon(57347)   on 2011-06-03 16:07:06 (Show Source): You can put this solution on YOUR website! Given the random variable X = {5, 6} with P(5) = 0.2 and P(6) = 0.8. Find E(X). ----------------- E(x) = (0.2)5 + (0.8)6 ------- E(x) = 1 + 4.8 ------- E(x) = 5.8 ============== Cheers, Stan H. ==============
 Probability-and-statistics/457091: The following distribution is not a probability distribution because The following distribution is NOT a probability distribution because X -2 -1 0 1 2 P(X) 0.13 0.19 0.42 0.16 0.27 A) the values of the variable are negative. B) the probability values are not increasing. C) the probability values do not add to 1. D) the probability values are not discrete. 1 solutions Answer 313657 by stanbon(57347)   on 2011-06-03 16:03:25 (Show Source): You can put this solution on YOUR website!The following distribution is NOT a probability distribution because X -2 -1 0 1 2 P(X) 0.13 0.19 0.42 0.16 0.27 ============================================= Ans: C ------------- Cheers, Stan H. =================== A) the values of the variable are negative. B) the probability values are not increasing. C) the probability values do not add to 1. D) the probability values are not discrete.
 Probability-and-statistics/457101: hello here is a exam question i have on my practice paper : Emilia picks up a chocolate at random from a box. All the chocolates in the box are identically wrapped. The probability that she picks a caramel is 5/8. Sophie picks a chocolate at random from a different box. All the chocolates in Sophie's box are identically wrapped. The probability that she picks up a caramel is denoted by P. The probability that both Emilia and Sophie pick a caramel is 1/4 Emilia's box P(emilia picks a caramal)=5/8 Sophies box P(Sophie picks a caramel)=P (a)Work out the value of P. (b)calculate the probability that Neither emilia nor sophie picks a caramel Please help me with these and tell me step step instructions what to do and tips thanks so much for your time 1 solutions Answer 313656 by stanbon(57347)   on 2011-06-03 16:00:16 (Show Source): You can put this solution on YOUR website! Emilia picks up a chocolate at random from a box. All the chocolates in the box are identically wrapped. The probability that she picks a caramel is 5/8. Sophie picks a chocolate at random from a different box. All the chocolates in Sophie's box are identically wrapped. The probability that she picks up a caramel is denoted by P. The probability that both Emilia and Sophie pick a caramel is 1/4 Emilia's box P(emilia picks a caramal)=5/8 -------------------- Sophies box P(Sophie picks a caramel)=P -------------------- Note: The events are independent. So, P(both pick caramel) = P(Emilia picks caramel)*P(Sophie picks caramel) ----- (a)Work out the value of P. Solve (5/8)P = (1/4) p = (1/4)/(5/8) P = 2/5 ======================== (b)calculate the probability that Neither emilia nor sophie picks a caramel P(Emelia does not pick caramel) = (3/8) P(Sophie does not pick caramel) = (3/5) P(neither picks are caramel) = (3/8)(3/5) = 9/40 =================================================== Cheers, Stan H.
 Probability-and-statistics/457104: Of the 6 courses offered by the music department at her college, Kay must choose exactly 2 of them. How many different combinations of 2 courses are possible for Kay if there are no restrictions on which 2 courses she can choose?1 solutions Answer 313652 by stanbon(57347)   on 2011-06-03 15:52:20 (Show Source): You can put this solution on YOUR website! Of the 6 courses offered by the music department at her college, Kay must choose exactly 2 of them. How many different combinations of 2 courses are possible for Kay if there are no restrictions on which 2 courses she can choose? --- Ans: 6C2 = (6*5)/(1*2) = 15 ============================= Cheers, Stan H.
 Probability-and-statistics/457118: he mean (μ) of the scale is 55 and the standard deviation (σ) is 14. Assuming that the scores are normally distributed, what is the PROBABILITY that a score falls below 39?1 solutions Answer 313650 by stanbon(57347)   on 2011-06-03 15:50:35 (Show Source): You can put this solution on YOUR website!he mean (μ) of the scale is 55 and the standard deviation (σ) is 14. Assuming that the scores are normally distributed, what is the PROBABILITY that a score falls below 39? ---------------- z(39) = (39-55)/14 = -1.1429 =================================== P(x < 39) = P(z < -1.1429) = 0.1265 ========== Cheers, Stan H.
 Probability-and-statistics/457120: Consider a large population of scores from a bimodal distribution with mean μ = 64 and standard deviation σ = 16. Imagine that you randomly selected a sample from that population that contained 47 scores. What is the PROBABILITY that the mean of the sample is above 70?1 solutions Answer 313649 by stanbon(57347)   on 2011-06-03 15:47:16 (Show Source): You can put this solution on YOUR website!Consider a large population of scores from a bimodal distribution with mean μ = 64 and standard deviation σ = 16. Imagine that you randomly selected a sample from that population that contained 47 scores. What is the PROBABILITY that the mean of the sample is above 70? ------------------- t(70) = (70-64)/[16/sqrt(47)] = 2.5709 ------ P(x-bar > 70) = P(t > 2.5709 when df 46) = 0.0067 =================================================== Cheers, Stan H.
 Rational-functions/457113: Could you please help me with this equation? Thanks!! 9ac/a^2-c^2 - a-c/a+c Subtract. Simplify by removing factors of 1.1 solutions Answer 313647 by stanbon(57347)   on 2011-06-03 15:16:33 (Show Source): You can put this solution on YOUR website!9ac/a^2-c^2 - a-c/a+c Subtract. Simplify by removing factors of 1. 9ac/a^2-c^2 - a-c/a+c Subtract. Simplify by removing factors of 1. ----------------------------------- Factor: (9ac)/[(a-c)(a+c)] - (a-c)/(a+c) ----- lcd = (a-c)(a+c) ---- Rewrite each fraction with the lcd as its deonominator: (9ac)/lcd - [(a-c)(a-c)]/lcd --------------- Combine the numerators over the lcd: [9ac-(a^2-2ac+c^2)]/lcd ----- Simplify: ----- (-a^2+11ac-c^2)/lcd --------------- = [-a^2+11ac-c^2)/[a^2-c^2] ------------- Cheers, Stan H. =============
 Probability-and-statistics/457114: According to popular belief, 80% of adults enjoy drinking beer. Choose a group of 2 adults at random. The probability that all of them enjoy drinking beer is: A) 0.500 B) 1.600 C) 0.400 D) 0.640 1 solutions Answer 313646 by stanbon(57347)   on 2011-06-03 15:05:39 (Show Source): You can put this solution on YOUR website!According to popular belief, 80% of adults enjoy drinking beer. Choose a group of 2 adults at random. The probability that all of them enjoy drinking beer is: --------- ---- The events are independent: P(enjoy and enjoy) = P(enjoy)^2 = 0.8^2 = 0.64 ================== Cheers, Stan H. ================== A) 0.500 B) 1.600 C) 0.400 D) 0.640
 Probability-and-statistics/456928: Could you please help me to solve this part question. Must show work. A population has a normal distribution with a mean of 130 and a standard deviation of 30. Find the probability that a single element selected from the population will have a value between 139.50 and 167.25. Use the population described in problem 2 above. Now find the probability that the sample mean for a sample of 16 elements selected from the population will be between 139.50 and 167.25. Explain the reason the answers to the above are different. 1 solutions Answer 313592 by stanbon(57347)   on 2011-06-03 09:36:51 (Show Source): You can put this solution on YOUR website!A population has a normal distribution with a mean of 130 and a standard deviation of 30. Find the probability that a single element selected from the population will have a value between 139.50 and 167.25. --- z(139.5)=(139.5-130)/30 = 0.3167 z(167.25)=(167.25-130)/30 = 1.2417 --- P(139.5< x < 167.25) = P(0.3167< z <1.2417) = 0.2686 ============================================================ Use the population described in problem 2 above. Now find the probability that the sample mean for a sample of 16 elements selected from the population will be between 139.50 and 167.25. ------- t(139.5)=(139.5-130)/[30/sqrt(16)] = 1.2667 t(167.25)=(167.25-130)/[30/sqrt(16)] = 4.9667 ============= P(139.5 < x-bar < 167.25) = P(1.2667 < t < 4.9667 when df = 15) = 0.1122 ========================================================= Cheers, Stan H. =============
 Probability-and-statistics/456931: Could you please help me to solve this question Must show work A sample of 100 elements produced a mean of 84 and a standard deviation of 8. Find the 99% confidence interval for the population mean. Assume the population is normally distributed.1 solutions Answer 313591 by stanbon(57347)   on 2011-06-03 09:23:14 (Show Source): You can put this solution on YOUR website!A sample of 100 elements produced a mean of 84 and a standard deviation of 8. Find the 99% confidence interval for the population mean. Assume the population is normally distributed. ---- sample mean = 84 Margin of Error = ME = 2.5758*8/sqrt(100) = 2.0607 ------------------------------ 99% CI: 84-2.0607 < u < 84+2.0607 ---- 99% CI: 81.94 < u < 86.06 ============================================ Cheers, Stan H. ===========
 Probability-and-statistics/457037: A coin-operated coffee machine made by BIG Corporation was designed to discharge a mean of eight ounces of coffee per cup. If it dispenses more that that on average, the corporation may lose money, and if it dispenses less, the customers may complain. BIG Corporation would like to estimate the mean amount of coffee, μ, dispensed per cup by this machine. BIG will choose a random sample of cup amounts dispensed by this machine and use this sample to estimate μ. Assuming that the standard deviation of cup amounts dispensed by this machine is 0.35ounces, what is the minimum sample size needed in order for BIG to be 95%confident that its estimate is within 0.05ounces of μ? Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements). 1 solutions Answer 313590 by stanbon(57347)   on 2011-06-03 09:16:59 (Show Source): You can put this solution on YOUR website!A coin-operated coffee machine made by BIG Corporation was designed to discharge a mean of eight ounces of coffee per cup. If it dispenses more that that on average, the corporation may lose money, and if it dispenses less, the customers may complain. BIG Corporation would like to estimate the mean amount of coffee, μ, dispensed per cup by this machine. BIG will choose a random sample of cup amounts dispensed by this machine and use this sample to estimate μ. Assuming that the standard deviation of cup amounts dispensed by this machine is 0.35ounces, what is the minimum sample size needed in order for BIG to be 95%confident that its estimate is within 0.05 ounces of μ? ---------------------------------------- n = [zs/E]^2 -- z = 1.96 s = 0.35 E = 0.05 ---------- n = [1.96*0.35/0.05]^2 = 13.72^2 = 188.238 ~ 189 when rounded up ========================== Cheers, Stan H.
 Probability-and-statistics/456948: I'm having difficulty with several parts of this problem. I've made some headway, but I'm fearful that it is all somewhat off so far. I'm finding this very frustrating. It's a multi-part problem. I'm really trying to learn the reasons behind these questions. Would anyone be willing to help me with the steps of how to get to the conclusions? It might be supposed that rainy days would tend to be cooler than days without rain, other factors being equal. Furthermore, if that rain is widespread enough to affect more than one reporting station in a city, the effect might be expected to be even more pronounced. The following output from the Excel Analysis ToolPak, shows a regression analysis on a year of Houston weather data. Precipitation data were collected for George Bush International Airport (KIAH) and for Hobby Airport (KHOU), coded as 1 for precipitation and 0 for no precipitation. These values were used as "binary predictors". The dependent variable is the departure of the daily high temperature from the 30 year normal for the date. So, for example, a day with a high temperature two degrees below normal would be shown as a departure value of -2.0. Regression Statistics Multiple R 0.1768 R Square 0.0313 Adjusted R Square 0.0258 Standard Error 6.8608 Observations 360 ANOVA ______________df________SS____________MS________F_______Significance F Regression_____2________542.19_______271.10____5.7594____0.0035 Residual_______357______16804.16_____47.07 Total__________359______17346.35 _________Coefficients____Standard Error_t Stat__P-value_Lower 95%_Upper 95% Intercept______ 2.6668___0.4381________ 6.0878__0.0000__ 1.8053___3.5283 KIAH_Precip____-2.3840___1.0423________-2.2873__0.0228__-4.4337___-0.3342 KHOU_Precip____-0.6236___0.9994________-0.6240__0.5331__-2.5890___1.3419 a) Write the fitted regression equation b) What does the R-squared value tell you? c) What, if any, meaning does the intercept have? d) Other things being equal, is the temperature warmer on days with or without rain at KIAH? e) Does the 95% confidence interval for KIAH_Precip slope give us any confidence that precip vs. no precip at KIAH is an important predictor of temperature? f) Describe the fit of this regression 1 solutions Answer 313587 by stanbon(57347)   on 2011-06-03 09:09:48 (Show Source): You can put this solution on YOUR website!It might be supposed that rainy days would tend to be cooler than days without rain, other factors being equal. Furthermore, if that rain is widespread enough to affect more than one reporting station in a city, the effect might be expected to be even more pronounced. The following output from the Excel Analysis ToolPak, shows a regression analysis on a year of Houston weather data. Precipitation data were collected for George Bush International Airport (KIAH) and for Hobby Airport (KHOU), coded as 1 for precipitation and 0 for no precipitation. These values were used as "binary predictors". The dependent variable is the departure of the daily high temperature from the 30 year normal for the date. So, for example, a day with a high temperature two degrees below normal would be shown as a departure value of -2.0. Regression Statistics Multiple R 0.1768 R Square 0.0313 Adjusted R Square 0.0258 Standard Error 6.8608 Observations 360 ANOVA ______________df________SS____________MS________F_______Significance F Regression_____2________542.19_______271.10____5.7594____0.0035 Residual_______357______16804.16_____47.07 Total__________359______17346.35 Variables Coficents____SE------------t Stat__P-value_Lower 95%_Upper 95% Intercept-------2.6668___0.4381________ 6.0878__0.0000__ 1.8053___3.5283 KIAH_Precip____-2.3840___1.0423________-2.2873__0.0228__-4.4337___-0.3342 KHOU_Precip____-0.6236___0.9994________-0.6240__0.5331__-2.5890___1.3419 ============================================================================== a) Write the fitted regression equation y = 2.6668-2.3840(KIAH)-0.06236(KHOU) ------------------------------------------------- b) What does the R-squared value tell you? R^2 =0.0313 tells you that 3% of the variation in temperature is explained by the KIAH and KHOU predictors. ------------------------------------------------- c) What, if any, meaning does the intercept have? The daily high-temp difference from the 30-year norm is 2.6668 degrees if we don't consider KIAH and KHOU data. ------------------------------------------------ d) Other things being equal, is the temperature warmer on days with or without rain at KIAH? Because the coef for KIAH is negative, rain at KIAH indicates a lower temp. ------------------------------------------------ e) Does the 95% confidence interval for KIAH_Precip slope give us any confidence that precip vs. no precip at KIAH is an important predictor of temperature? Yes because the coef for KIAH is in the 95% CI. ------------------------------------------------ f) Describe the fit of this regression The fit is pretty weak as seen in the low R-value. ====================================================== Cheers, Stan H.
 Probability-and-statistics/456847: A machine used to fill half-gallon-sized milk containers is regulated so that the amount of milk dispensed has a mean of 64 ounces and a standard deviation of 0.11 ounces. You randomly select 40 containers and carefully measure the contents. The sample mean of the containers is 64.05 ounces Does the machine need to be reset? Explain your reasoning.1 solutions Answer 313475 by stanbon(57347)   on 2011-06-02 18:36:04 (Show Source): You can put this solution on YOUR website!A machine used to fill half-gallon-sized milk containers is regulated so that the amount of milk dispensed has a mean of 64 ounces and a standard deviation of 0.11 ounces. ------------- You randomly select 40 containers and carefully measure the contents. The sample mean of the containers is 64.05 ounces Does the machine need to be reset? Explain your reasoning. ------ Ho: u = 64 Ha: u is not 64 ------------------------ x-bar = 64.05 t(65.05) = (64.05-64)/[0.11/sqrt(40)] = 2.8748 ---------------------------- p-value = 2*P(t > 2.8748 when df=39) = 0.006516.. --- Since the p-value is less than 1%, reject Ho at the 1% level of significance. --- Test results support resetting the machine. =============================================== Cheers, Stan H.
 Probability-and-statistics/456836: A vending machine dispenses coffee into an eight-ounces cup. The amounts of coffee dispensed into the cup are normally distributed with a standard deviation 0.01 ounce. You can allow the cup to overfill 1% of the time. What amount you should set as the mean amount of coffee to be dispensed?1 solutions Answer 313471 by stanbon(57347)   on 2011-06-02 18:16:03 (Show Source): You can put this solution on YOUR website!A vending machine dispenses coffee into an eight-ounces cup. The amounts of coffee dispensed into the cup are normally distributed with a standard deviation 0.01 ounce. You can allow the cup to overfill 1% of the time. What amount you should set as the mean amount of coffee to be dispensed? --------------------------------------- Find the z-value with a right tail of 1%: z = 2.3263 ----------- Find the x-value using x = zs+u x = 2.3263*0.01+8 = 8.0233 ======== Let that be the mean amount of coffee to be dispensed. ======================================================= Cheers, Stan H.
 Probability-and-statistics/456839: In a survey of women in the United States (ages 20-29), the mean height is 69.9 inches with a standard deviation 3.0 inches. a. What height represents the 95th percentile? b. What height represents the first quartile? 1 solutions Answer 313467 by stanbon(57347)   on 2011-06-02 18:02:26 (Show Source): You can put this solution on YOUR website!In a survey of women in the United States (ages 20-29), the mean height is 69.9 inches with a standard deviation 3.0 inches. a. What height represents the 95th percentile? Find the z-value with a left tail of 95%: z = 1.645 Use x = zs+u x = 1.645*3+69.9 = 74.835 inches ---------------------------------------------------- b. What height represents the first quartile? Find the z-value with a left tail of 25%: z = -0.6745 x = -0.6745*3+69.9 = 67.877 inches ================================== Cheers, Stan H.
 Polynomials-and-rational-expressions/456838: Find the greatest common factor. x^7,x^31 solutions Answer 313465 by stanbon(57347)   on 2011-06-02 17:56:11 (Show Source): You can put this solution on YOUR website!Find the greatest common factor. x^7,x^3 --- Greatest COMMON factor is x^3
 Linear_Algebra/456831: In 1995, the life expectancy of males in a certain country was 73.8 years. In 2001, it was 77.1 years. Let E represent the life expectancy in year t and let t represent the number of years since 1995. The linear function E(t) thats fits the data is E(t)=1 solutions Answer 313464 by stanbon(57347)   on 2011-06-02 17:52:28 (Show Source): You can put this solution on YOUR website!In 1995, the life expectancy of males in a certain country was 73.8 years. In 2001, it was 77.1 years. Let E represent the life expectancy in year t and let t represent the number of years since 1995. --------------- Two points: (0,73.8), (6,77.1) --------------- intercept = 73.8 slope = (77.1-73.8)/6 = 0.55 ------- The linear function E(t) thats fits the data is E(t)= 0.55t+73.8 =================================================================== Cheers, Stan H.
 Inequalities/456832: what is the answer to this? n/5 > - 9/5 1 solutions Answer 313462 by stanbon(57347)   on 2011-06-02 17:47:27 (Show Source): You can put this solution on YOUR website!n/5 > - 9/5 ----- Multiply both sides by 5 to get: n > -9 ============ Cheers, Stan H.
 Average/456801: Can you please explain to me how to calculate Fahrenheit in a word problem. I have looked on the internet but it does not give a full explanation how to resolve a question that involves temperatures? 1 solutions Answer 313444 by stanbon(57347)   on 2011-06-02 16:39:26 (Show Source): You can put this solution on YOUR website!Can you please explain to me how to calculate Fahrenheit in a word problem. I have looked on the internet but it does not give a full explanation how to resolve a question that involves temperatures? ----- Freezing temperatures: Celcius = 0, Farenheight = 32 Boiling temperatures: Celcius = 100, Farenheight = 212 ----- Generate two Equations: C to F: slope = (212-32)/(100-0) = 180/100 = 9/5 ---- intercept: (0,32) --- Equation: F = (9/5)C + 32 ======================= Solve for "C": F-32 = (9/5)C C = (5/9)(F-32) ========================= Those are the 2 conversion formulas. ========================================= Cheers, Stan H.
 Linear-equations/456787: solve the system by the substitution method 2x-3y=11 x-4y=01 solutions Answer 313443 by stanbon(57347)   on 2011-06-02 16:32:15 (Show Source): You can put this solution on YOUR website!solve the system by the substitution method 2x-3y=11 x-4y=0 ----------------- Solve for "x": x = 4y ---- Substitute for "x" and solve for "y": 2(4y)-3y = 11 8y-3y = 11 5y = 11 y = 11/5 ------------ x = 4y = 4(11/5) = 44/5 ------------------------------ Cheers, Stan H.