請問第1題和第2題怎麼算呢？不太懂什麼時候需要使用自由度(df)

以後練習 ✓ 1. 100 位民眾之舒張壓平均數為73 mmHg,標準差s = 11.6 mmHg, 試計算p之95%CI及99% CI估計值是多少? ↓ 2. 100 筆地下水質監測資料,含As之平均濃度為0.15 mg/L,標準差 S= 0.2 mg/L,試計算 之 95% CI及 99% CI? 3. 若 38 名男性, = 74.9,S = 12;若 45 名女性, = 71.8,S, = 11;計算 - 之 95% CI 估計值是多少? 4. 若採樣 100 筆台中地區空氣中 PM.s濃度之X = 65 ug/m²,S = 10 ug/m²;若採樣 120 筆南投埔里地區空氣中PM 濃度之 = 80 ug/ m²,Sz = 15 ug/m²,請計算台中地區與埔里地區之 PM; 濃度差的 95% CI 估計值是多少? ading on aod teem

以下是心統的題幹與問題 想請問第以下兩題是否有計算錯誤？

1. Any of you who have listened to children tell stories will recognize that, unlike adults, they tend to recall stories as a sequence of actions rather than as an overall plot. Their descriptions of a movie are filled with the phrase "and then...." An experimenter with supreme patience asked 50 children to tell her about a given movie. Among other variables, she counted the number of "and then..." statements, which is the dependent variable. The data follow: 18 15 22 19 18 17 18 20 17 12 16 16 17 21 23 18 20 21 20 20 15 18 17 19 20 23 22 10 17 19 19 21 20 18 18 24 11 19 31 16 17 15 19 20 18 18 40 18 19 16 簡單翻譯:心理學家請小孩描述電影情節,並記錄他們使用「然後」的次數,50 位小孩的次數 如上。

算數平均數 中位數 眾數 89 90 91題 求解 算不出來😭 有試著算過 可以順便告訴我我哪個步驟出問題了 謝謝🙏

: (B) 已分組找組中美 89 【單選題】 某班學生之成績次數表如下所示。試求學生們成績之算術平均數? <法> A 本組次數 組中美 (A) 64.744 (B) 59.338 (C) 55.129 -3 (D) 70.961 答案:(A) 30-39 40-49 50-59 160-69 70-79 80-89 90-99 合計 2 2 5 95 (D) 68.1 答案: (C) -30 (D) 64.5 答案: (D) 4-5 1-20 2 41+1 2 8 5 -10 21 5 平移 90. 【單選題】 承上題,試求學生們成績之中位數? (A) 58.3 ① (B) 60.9 (C) 64.5 zl 7 (65) 5 10 160~691 O 85 20 30. (3)x2+(-2)x3+(-1)x8+①+1x7+2X4+3x2 41 35x2+11+95x2 41 3 平移縮小法→每个45 除10 14th = 60分 8 * 21th ² y → y = 60+ 29th=69157 比列 距离 91. 【單選題】 承上題,試求學生們成績之眾數約為? (A) 56.5 (B) 58.0 (C) 60.1 365243 7 8 比列 x10=65 + 4x10=65: 縮小 x 10 = 68.75

各位請問這兩題怎麽算，謝謝

測驗資訊 作答區 填充 配分:[10.00] 題數 8/10 1. (1) Evaluate the function f(x) = x² - (2²/1000) for x = 1,0.8, 0.6, 0.4, 0.2, 0.1, 0.05. and guess lim | 7² 2 lim 1-0 (2) Evaluate the function f(x) for x = 0.04, 0.02, 0.01, 0.005, 0.003, 0.001 2¹ 1000 2 頁數 8/10 1000 t 剩下時間 2:49:34 and guess. 送出管家,湖上頁 送出答案,超下克 ※※送〔為了確保您的答案能完整記錄,作答完畢後,請務必點選「送出答案,結束測驗』繳卷】※※※ ※※※【注意:點選視窗右上方"X"「只關閉視窗,不儲存答案』】※※※

請問紅色圈起來的地方是為什麼？ 謝謝🙏

7−(−2+h)— √7−(−2) h −1 −1 √9-h+3 V9+3 My Ax As h→0. = 9-h-3 h 6 = 9-h-3√9-h+3 (9-h)-9 -1 = √9+h+3h(√√9+h+3)¯¯ √√9+h+3² Th ⇒ at P(-2,3) the slope is =

想問第42題

25. lim 8118 27. lim X18 A 29. lim 31. lim 33. lim √x - √√x x→→∞ √√√x + √√x 35. lim 2x²/3 - x1/3 + 7 xxx x8/5 + 3x + √x 37. lim x→∞ x + 1 39. lim x - 3 x→∞ √4x² + 25 1 x→0+ 3x 1-x² 2√x + x¹ 3x - 7 41. lim 1. 2 X-2 X- 43. lim √x² + 1 45. a. lim 3 2x x-8+ x + 8 + 7x 4 x→7 (x-7)² 2 2 46. a. lim x-0+ x1/5 4 2 x-0+ 3x¹/3 B Infinite Limits Find the limits in Exercises 37-48. Write ∞ or -∞ where appropriate. 5 x →0-2x x - 3x 26. lim√√³+x-2 b. lim 2 + √x x→∞ 2 - √x 28. lim 30. lim x¹ + x4 x-xxx-²-x-3 32. lim Vì - 5x + 3 x-∞ 2x + x2/3 - 4 34. lim 36. lim 2 x→0 3x¹/3 xxx +1 2 b. lim x→0 x1/5 4 - 3x3 x-∞ √6 +9 38. lim 40. lim √x² + 1 42. lim x-3x - 3 3x x-5-2x + 10 44. lim -1 x→0 x²(x + 1) 19 lime 1

請問大一統計學這兩題有人看得懂嗎🥲

2. To study the attitude of mother and father towards their daughter participating in the outdoor games, a sample of 200 fathers and 250 mothers was investigated. In these two samples, 120 fathers and 50 mothers preferred their daughter to go for outdoor activities. Can it be concluded at a=0.01 level of significance that the proportion of father is significantly higher than that of mother in preferring their daughters for outdoor activities? 3. A firm has a generous but rather complicated policy concerning end-of-year bonuses for its lower-level managerial personnel. The policy's key factor is a subjective judgment of "contribution to corporate goals." A personnel officer took samples of 24 female and 36 male managers to see whether there was any difference in bonuses, expressed as a percentage of yearly salary. The data are listed here: Gender Bonus Percentage Female 9.2, 7.7, 11.9, 6.2, 9.0, 8.4, 6.9, 7.6, 7.4, 8.0, 9.9, 6.7, 8.4, 9.3, 9.1, 8.7, 9.2, 9.1, 8.4, 9.6, 7.7, 9.0, 9.0, 8.4 10.4, 8.9, 11.7, 12.0, 8.7, 9.4, 9.8, 9.0, 9.2, 9.7, 9.1, 8.8, 7.9, 9.9, 10.0, 10.1, 9.0, 11.4, 8.7, 9.6, 9.2, 9.7, 8.9, 9.2, 9.4, 9.7, 8.9, 9.3, 10.4, 11.9, 9.0, 12.0, 9.6, 9.2, 9.9, 9.0 (1) Is there significant evidence that the mean bonus percentage for males is larger than the mean bonus percentage for females? Use a a=0.05. (2) Estimate the difference in the mean bonus percentages for males and females using a 95% confidence interval. Male

想詢問這個該如何證明？

of duration relat- ing to the maturity, yield, and coupon interest of the security being analyzed. Duration and Maturity A comparison of Tables 9-6, 9-3, and 9-7 indicates that duration increases with the maturity of a fixed-income asset or liability, but at a decreasing rate:4 OD OM > 0 d²D dM² <0 To see this, look at Figure 9-4, where we plot duration against maturity for a three-year, a two-year, and a one-year U.S. Treasury bond using the same yield of 4 This is the case for the vast majority of securities. It needs to be noted, however, that for bonds selling below par, duration increases at a decreasing rate up to a point. At long maturities (e.g., 50 years) dura- tion starts to decline. Few bonds in the market have a maturity long enough to see this decline.

求解6.（a) 雖然有算出答案，但不確定過程、答案有沒有錯 算式在第二張照片中

(a) 試求上題 (a) 的 cd.f. Fx(x) (5%) P(X-10+ 3 (b).求出上題(b) 的下列之事件機率: 1or 1 = = * = + 4x(+)) P(X=1 or 3) . (5%) (a) x1 4*4*(1+(+)) P(1<X<4)。(5%) (iii)P(X>5) (5%) = 11 x 15 11 16 x 16 255 256 # (b) te-o — 。 Fx (x) 1 10. 14 6 éx X 3 3 6 105 105 1 (b) f(x) = -e-x/304 0 ≤ x < (5/ ce³+ + -e ³0 = | c²1-e 30 30 x te 。 2 4 10 105 5. 針對下列各函數,求常數c 使得f (x)是一隨機變數 X 的 p.d.f. : (a) f(x) = cvx, 0 ≤ x ≤ 4°(5%) (a) c Jo + C+ C2+C15+C[41 (3C+FC+8c=1 C= C =- = | 設化是抽到瑕疵電腦的數 X是抽2臺電腦抽到瑕疵品的機率 C5 10 5 if x=03 28 5 6 1 15 21 28 105 105 105 8 36 3 Cx-(+) 15 16 Cxlim 15 19 3+√2+√3 9-(5+256) 66 18 105 105 105105105 9|10|11|12 55 (eº + e¯ 30 + ₁ + e 30+) = | C = - 二 =1 6. 續上題 (b) 1-6-30 1.030 (a) 試求累積分布函數(cumulative distribution function(x(x) x≤x)) (5%) \C: (b) P(15 ≤ X ≤ 30)。(5%) C = 848 O 7.已知8台電腦中有3台是瑕疵品。今隨機從此8台電腦中抽取2台,請問所抽取電 疵品個數之機率質量函數(p.m.f.) 為何?(7.5%) (29(x)=75x= it x = ²₂ ( 1 = 28 (Px (x) = 18, X = x=2, 15 285 LPx(x)=375X 4+. 16- 4个 143 X=

各位大神 求解 拜託🙏🆘

CA 梁柱 一截面積是原來的兩倍 22. Two girders are made of the same material. Girder A is twice as long as girder B and Bo DE C B³ has a cross-sectional area that is twice as great. The ratio of the mass density of girder A to the mass density of girder B is: 質量密度 A) 4 B) 2 C) 1 D) 1/2 E) 1/4 29. During a short interval of time the speed v in m/s of an automobile is given by v=at² + bt³, where the time t is in seconds. The units of a and b are respectively: A) m-s²; m-s4 s³/m; s4/m dv=at+bt² B) C) m/s²; m/s³ D) m/s³; m/s4 E) m/s4; m/s5 34. The position y of a particle moving along the y axis depends on the time t according to the equation B) C) D) E) 質固定,体2倍 #%# 40. The coordinate of an object is given as a function of time by x = 7t-31², where x is in meters and t is in seconds. Its average velocity over the interval from t = 0 to t = 2 s is: A) 5 m/s X₂=0, X₁₂=14-12=2 B) -5 m/s 11 m/s C) D) -11 m/s (2-0)/2=1 E) -14.5 m/s 41. The coordinate of a particle in meters is given by x(t) = 16t-3.01³, where the time t is in seconds. The particle is momentarily at rest at t = A) 0.75 s A y = at -bt². The dimensions of the quantities a and b are respectively: A) L²/T, L³/T2 B) L/T², L²/T C) L/T, L/T² D) L³/T, T²/L E) none of these 1.3 s 5.3 s 7.3 s 9.3 s V(t) = 16-9+² =- [9t²-16) =-(3t+4) 13t-4) =+= 0.75 45. The velocity of an object is given as a function of time by v = 4t-312, where v is in m/s and t is in seconds. Its average velocity over the interval from t=0 to t=2 s: A) is 0 B) is -2 m/s C) is 2 m/s D) is -4 m/s E) cannot be calculated unless the initial position is given 49. A particle moves along the x axis according to the equation x = 61² where x is in meters and t is in seconds. Therefore: E B) C) D) E) A) the acceleration of the particle is 6 m/s² t cannot be negative the particle follows a parabolic path each second the velocity of the particle changes by 9.8 m/s none of the above