求解謝謝🙇🏻‍♀️🙇🏻‍♀️只要一小題就好

想請問這題給的答案是對的嗎？因為題目裡有寫明security X跟security Y的權重分別是70%跟30%，但是到了答案裡又變成了60%跟40%，有點困惑🥲🙏

收益率 A Ecr × (70%) 14% Y (30%) 10% 4. You are considering investing $1,000 in a complete portfolio. The complete portfolio is composed of Treasury bills that pay 5% and a risky portfolio, (P, constructed with two risky securities, X and Y. The optimal weights of X and Y in P are 70% and 30%, respectively. X has an expected rate of return of 14%, and Y has an expected rate of return of 10%. The dollar values of your positions in X, Y, and Treasury bills would be and , respectively, if you decide to hold a complete portfolio that has an expected return of 8%. Treasury the $110 $1000 profoliots a f .08(1000)=F(.05)+(1000 - F)[.6(.14)+.4(.10)] F = 595 X = (1,000-595)(.6)=243 Y = (1,000-595)(.4)=162 7-175% P X-1 14% Y- 10% □

這兩題該怎麼寫

5. Suppose that a reaction C2H6C2H4+H2 is believed to occur by the mechanism: 1. C2H6-2CH3 2. CH3+ C2H6 CH4+ C2H5 ->> 3. C2H5 C2H4+H 4. C2H6+H→C2H5+H2 5. 2 C2H5 C4H10 . Apply the steady-state treatment to C2H5 CH3 Hand obtain an expression for the overall rate. 6. The reaction between chlorine and chloroform in the gas phase follows the stoichiometric equation CHC13 C12 CCl4 + HCl It is believed to occur by the mechanism Cl₂+hv 2C1 Cl+CHCl3 HCl + CC13 CC13 C12 CCl4 + Cl 2C1→Cl₂ Assume that the steady-state treatment can be applied to C1,CC13 and obtain an expression for the overall rate.

求解🙇 課本給的答案是(a)1.91cm/s(b)left

41 M In Fig. 28.31, a metal wire of mass m = 24.1 mg can slide with negligible friction on two horizontal parallel rails separated by distance d=2.56 cm. The track lies in a vertical uniform mag- netic field of magnitude 56.3 mT. At time t = 0, device G is con- nected to the rails, producing a constant current i = 35.0 mA in the wire and rails (even as the wire moves). At t = 61.1 ms, what are the wire's (a) speed and (b) direction of motion (left or right)? d m B FIGURE 28.31 Problem 41.

113年地方特考 三等統計學 第四題 請問： 第二張圖片內 拒絕域的F值要怎麼算出來 這份考卷只有附Z分配、t分配表 並未附F分配的表

使用了三種肥料後農作物的生長高度(cm),如下表所示: B A C 20 27 21 27 24 17 26 21 20 24 22 25 「請建立ANOVA表,並在顯著水準0.1下檢定三種肥類對農作物的平均生長高度 是否不同? A B C 加總 20 2 21 68 27 24 17 68 26 20 21 67 24 25 22 71 加總 97 97 80 274 [EXG=6366 SST=6366 - 2742 = 109.667 12 SSTR = 97²+ 97² + 80² 274² = 48.167 4 4 - 4 12 SSE = SST - SSTR = 61.5 來源 SS df MS F 組間 48.167 2 24.084 F:3.525 組內 61.5 9 6.833 總變異 109.667 "

電路學 求解這兩題

P5.39 47 k2 w 220 k2 w 6V 6 V 時間:102 33 kQ 1k2 -6V -6 V g k2 5.40 圖P5.40 電路中的兩個運算放大器均為理 想。計算和202。 P5.40 15 V 15 Vo 春操作 Vol 400 -15 V 5002 (3. 15 V 2 kN 8 V 7 100230K2 300 運算放 -15 V :4002 5.41 2 4kQ 8 V + 5.42 圖 P5.41 陰影區內的電路對有限範圍的而 來說是一個定電流源。 (2) (b k2時,試求大小。 (a)所得時,RL 的最大 明電路的操作情 『操作條件下 3201 5.7 5.43 5.44

問一下這題

15. (電話響聲)當你打電話時,會響2 秒再靜音4秒,又再響2秒等等。以 下的圖形為打電話的人在電話響了 3 次掛掉的聲度 (sound-level)函數 S, 圖中 4 為房間中電話不響時的聲度, B則為電話響的聲度。 聲度 B y=S(t) A+ 2 4 6 8 10 12 14 時間(秒) mil 電話的三次響聲 (a) 求lim S(t) 1→4 (b) 求limS(t)。 1→7 考慮如下的極限, lim S(t) 1→6 這個極限並不存在,若由小於 6 (即由左邊 )來趨近,所得的 S(t) 值與由大於6(即由右邊)來趨近 所得值是不同。在下兩節,單邊極限 (第2.2節)和連續性(第2.3節) 會完整討論這種情形。

有人可以幫我嗎🥺🥺

18. (III) An automobile traveling 95 km/h overtakes a 1.10-km- long train traveling in the same direction on a track parallel to the road. If the train's speed is 75 km/h, how long does it take the car to pass it, and how far will the car have traveled in this time? See Fig. 2-39. What are the results if the car and train are traveling in opposite directions? v = 95 km/h 1.10 km v = 75 km/h FIGURE 2-39 Problem 18.

這是有機合成的題目，上課聽不太懂，求求以下三題🥲

Problems 4. Synthesis. Outline a retrosynthetic scheme for each of the following target molecules using the indicated starting material. Show (1) the analysis (including FGI, synthons, synthetic equivalents) and (2) the synthesis of each TM a. CI OMe c. Me CH₂OH d. OH O Br Br CI LOH OMe

想問這題：為什麼微分後一直算不出分母為729的分數？求解析🙏🙏謝謝！！

16% inal profip=500 1 + P 500(1 50 + t2 65. Population Growth A population of bacteria is ollar introduced into a culture. The number of bacteria P can be modeled by fut +50 50+42 4t 0+ 500 (4(50+1) 941-2+ (50+2) 62 sional 46 500 X- ≥31.55 resuls where t is the time (in hours). Find the rate of change of the population at t = 2. W (d) 729.