請問這兩題怎麼解 謝謝

Table 17-8 Mc=4 er. As For a certain small town, the table shows the demand schedule for water. Assume the marginal cost of supplying water is constant at $4 per bottle and there are no other costs. Price Quantity (bottles) TR R MR02 $9 200 1800 9 $8 400 $7 600 42 $6 800 3,200 & 800 8 $5 1000 5000 5 $4 1200 4800 4 $3 1400 $2 1600 4,200 3 3200/2 a monolopy 245 FX MR = MC 28 Refer to Table 17-8. If there were only one supplier of water, what would be the price and quantity? d. The price would be $7 per gallon and the quantity would be 600 gallons. b. The price would be $6 per gallon and the quantity would be 800 gallons. c. The price would be $5 per gallon and the quantity would be 1000 gallons. d. The price would be $4 per gallon and the quantity would be 1200 gallons.

請問這兩題怎算 感恩！我網路上找不到詳解…

經濟學 Ch18 尚 2019ch18 Table 17-36 The information in the table shows the total demand for water service in Takoma. Assume that there are two companies operating in Takoma. Each company that provides these services incurs an annual fixed cost of $400 and that the marginal cost of providing the service to each customer is exactly $2.00. Figures listed are for an annual service contract. FC = 400, MC = 2 Quantity Price 1R MR 0 60 0 0 100 55 5500 55 200 50 10000 50 300 45 13500 45 400 40 16000 40. 500 35 17500 35 600 30 18000 30 700 25 17 500 25 800 20 1600 2 900 15 13500 15 1000 10 10000 1100 5 1200 0 5500 0 12 Refer to Table 17-36. Assume there are two profit-maximizing water service providers in this market who had formed a successful cartel. Now assume that the cartel breaks down, so that they are not able to 假設卡特爾瓦解(有仅勢策略、

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

程式錯誤 請問該如何修改？

← → C CO 資結構構 ☆ !!! Ő E O colab.research.google.com/drive/1dNYtPmtBVPcvl2ZkcmoT5h8 檔案 編輯 檢視畫面 插入 執行階段 工具 說明 已儲存所有變更 + 程式碼 +文字 IN 1817 IndexError: list index out of range STACK OVERFLOW from traitlets. config. application import T from string import Template def MatrixMultiply (arrA, arrB, arrC, M, N, P): global C if M<=0 or N<=0 or P<=0: print([錯誤:維數]M, N, P必須大於0') return for i in range (M) : for j in range (P): Temp=0 for k in range (N) : Temp-Temp+int (arrA[i*N+k]*arrb [k*P+j]) arrC[i*P+j]-Temp print("請輸入矩陣A維數(M, N) : " ) M-int (input('M= ')) N=int (input (N= ')) A=[None] *M*N print([請輸入矩陣A的各個元素]) for i in range (M): for j in range (N) : A[i*N+j]=input('a%d%d=' % (i, j)) print(請輸入矩陣B維數(N, P);') N=int (input ('N= ')) P=int (input('P= ')) B=[None]*N*P | print([請輸入矩陣B的各個元素]') for i in range (N) : for j in range (P): B[i*P+j]=input('b%d%d=' % (1, j)) C 在這裡輸入文字來搜尋 PAMPLING 3241

想請問這題是如何解出來的？ 手寫的圖是我嘗試解的過程，但沒有解出答案，麻煩大家了🫠

21. A metal ion in a weak-field octahedral complex has two more unpaired electrons than the same ion does in a strong-field octahedral complex. Which of the following could the metal ion be? (A) Cr³+ (B)Mn²+ (C)Co²+ (D)Ti²+ (E)Cu²+

請問這題的Y hat=.22X+4.75 這個算式 是怎麼算出來的🥹🥹

r-Izxzy N SY b= xxx 5x a=y-bx 有八位學生的成績如下表,試計算其 SSt、SSreg、SSres 再根據所計算的結果求 決定係數及疏離係數。 f = .22X + 4.75 r² = SSreg SSt [學生 測驗一X 測驗二Y = A|B|C|D|E F G H 6 5 3 7 9 7 5 4 1.73 18 9+7+5+4+5+6+7+5 8 Y = = 學生 | A|B|C|D|E|F|G|H| |預測分數Ÿ 6.07 5.85 5.41 6.29 5.63 6.51 6.73 5.41 SSt = Σ(Y-V)² = 9 +1 +1 + 4 + 1 + 0 + 1 + 1 = 18 = SStres = (Y - Y) = 8.58 +1.32 +0.17+5.24+0.40+0.26 + 0.07 + 0.17 = 16.21 SSreg: Σ(Y− 1)² = 0.00 +0.02 +0.35 +0.08 +0.14 +0.26 +0.53 +0.35 = 1.73 = V0.901 = 0.95 |SSres SSt V1-r2: = 0.096 = =6 4 8 - 5 |16.21 18 6 397 3 5

！！求問！！ 統計學此題該如何解題

3.5 以下資料為減重班之成員,完成某指定運動所使用時間,時間單位以分計算 80 31 39 45 x6 x 16 35 37 25 42 32 58 19 16 56 1 34 36 12 48 38 37397 39 17 31 56 28 ¥ º 80 10 21 38 40 82 27 37 在此同時,也測量成員的心跳速率 63 89 75 80 74 65 90 85 92 84743998 91 87 26 82 90 93 77 74 89 85 91 102 69 87 96 83 72 92 88 85 68 78 73 86 85 92 90 分別以上述變數製作次數分配表,表格須包括原始次數,累積次數,和百分比 累次百分比。這些組距之組中點各為何?

普物下電學 想請問一下這兩題怎麼寫 沒有很懂題目在表達什麼 謝謝！

< EUR)- 2:30 10 4 Awes R 一個點電荷Q帶有多少電位能?要回答這個問題,我們可以先算出均勻帶電球(Example 22.9)的總 電位能,再將球半徑 R 趨近於零。 (a) 計算均勻帶電球的總電位能。 (b)推論一個點電荷Q帶有多少電位能- E- Or Amepl 91 90 HW5 P 47 3 附錄: (a)小題的兩種算法算法,可任選一方法計算- -R³ - Spherical insulator -Gassian surface Method 1. 假設一開始沒電荷,一次搬 dQ 的電量, 從無窮遠移過來,類似下圖四顆點電荷的例子, U=U₁+U₂+U₂ ~~ dU =Vdq = Method 2. dq=p4nr²dr = 10 4me7 47 Q₁.³ q(r)=p² -7.3 3 R³ q(r) Q-² V(r)= Amer 4eR = 假設電荷是一圈一圈的從無窮遠處移過來,列式如下, Q 30-² R³ dr = Q₁² 30₁.² 4лε R³ R³ 基本資訊 4 1 9091 +9092 +9093 +9192 +9193 +9293 Pol 102 103 112 113 123 1 -Σ² 4ME。 4.80 ky Py Ty ·dr =... (9.9元 II 4G 196 假設一開始電荷已經就位,計算所有電荷的電位能再除2,類似下圖四顆點電荷的例子, U = = Σ9V₁ 1-0 1 -9 寫作業

求解！如果選擇接受合作 該怎麼算合作後的利潤增長？ 會計科目不是很了解！

□ 1/3 博溫腳踏車公司 1983 年 5 月,博溫公司的行銷部門負責人李副總經理正在思索前天高檔公司採 購梁經理的提議。高檔公司是美國西北部一家平價百貨公司,目前成長快速,開始使 用自有品牌的商品銷售,梁先生是高檔公司運動器材部門的採購經理,他希望博溫公 司替高檔公司做OEM之腳踏車,高檔公司計畫將博溫公司生產的腳踏車取名為『挑 戰者』。 博溫公司已經有四十年歷史,在 1983 年,公司生產十種腳踏車,從小孩的訓練 車到成人的十二段變速跑車,每年的銷售額約一千萬美元,財務報表見表一。博溫公 司的產品銷售,主要是透過運動器材店、以及腳踏車專賣店,從沒有透過百貨公司經 銷。博溫公司的腳踏車品質與價格,都在平均水準以上(中高級產品),但卻不是最 高級產品。美國自行車市場區隔,詳見附錄一。 資產 現金 應收帳款 存貨 廠房設備淨值 資產總額 表一、博溫公司財務報表 資產負債表 1982 年 12 月 31日 銷售 銷貨成本 銷貨毛利 銷管費用 稅前淨利 所得稅 稅後淨利 $342 1,359 2,756 3.635 $8,092 -1- 負債與股東權益 應付帳款 應付費用 短期負債 長期負債 損益表 1982 年 股東權益 負債與股東權益總額 個案 (千美元) $10,872 8,045 2,827 2.354 473 218 - $255 (千美元) $512 340 2,626 1.512 4,990 3.102 $8.092

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