(Depreciation Computations (Depreciation Computations Solutio.docx

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This document discusses four methods for calculating depreciation: (1) straight-line, (2) units of production, (3) number of hours worked, and (4) sum of years digits. It provides examples of how to calculate depreciation under each method using information about an asset's cost, salvage value, expected production/hours, and useful life. It also shows the calculation of depreciation expenses over 10 years using the declining balance method.

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(Depreciation Computations
(Depreciation Computations
Solution
A) Depreciation under straight line method =( Cost - Salvage value)/ Number of years
= (315000-15000) / 10 i.e 30000
B) Depreciation under units of production method = Cost / Units expected to be produced *
Number of units produced
= 315000/240000 *25500 i.e 33469
C) Depreciation under numbers of hours worked bais = Cost / Expected total hours * Number of
hours worked
= 315000/25000*2650 i.e 33390
D) Sum of years digits = Depreciable base * Remaining life / sum of years
Depreciable base = Cost - Salvage value
= 315000-15000 i.e 300000
Sum of years = n(n+1)/2
= (10*11)/2 i.e 55
Calculation of depreciation under declining balance method
Depreciable amount = 300000
Annual rate = 20%

Year 1 = 315000*20%*2 i.e 126000
Year 2 = ((315000-126000)*20%*2 i.e 75600
Year 3 = (315000-126000-75600)*20%*2 i.e 45360
Years Depreciable Base Remaining life Depreciation factor Depreciation expense
1 300000 10 10/55 54545.45
2 300000 9 9/55 49090.90
3 300000 8 8/55 43636.36
4 300000 7 7/55 38181.81
5 300000 6 6/55 32727.27
6 300000 5 5/55 27272.72
7 300000 4 4/55 21818.18
8 300000 3 3/55 16363.63
9 300000 2 2/55 10909.09
10 300000 1 1/55 5454.54

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What is the backbone for the Internet? Briefly describe how the Internet works. Solution An internet backbone refers to one of the principal data routes betweeen large, strategically interconnected networks and core routers on the internet they provide a consistent way for \"internet service provider\"(isp) to keep and maintain data online with security. Internet: it is a globalsystem of interconnected physicl computer networks such as lan\'s,wan\'s and man\'s. in simple terms network is a network\'s of million\'s of public,private connected with the help of widely available internet working devices. Working of internet: The client initates the process and send request for particular internet service The n/w devices such as modem and ethernet card process the signal so that it can be sent over the local loop carrier or the signal carriers. Loop connects the loaction of user to isp point of presence. it accepts connection from users authentic connetcions.the signal are then transfered to isp regional network consisting of interconnected routers in different cities served by isp. If the packet is supposed to reach the host, whichh is directly served by the isp, then it is delivered to the host.otherwise the packet is sent to isp backbone operator if the packet recived by the backbone is supposed to reach a server then the packet is forwarded to the nearest router. .

What is the backbone for the Internet- Briefly describe how the Intern.docx

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What is the pH of a 0.1 M aqueous solution of NaH2PO4 Solution NaH2PO4 --------------> Na^+ (aq) + H2Po4^- (aq) Ka = 6.8*10^-8 Kb = Kw/Ka = 1*10^-14/6.8*10^-8 = 1.47*10^-7 H2PO4^- (aq) + H2O(l) ----------------> H3Po4(aq) + OH^- (aq) I 0.1 0 0 C -x +x +x E 0.1-x +x +x Kb = [H3PO4][OH^-]/[H2Po4^-] 1.47*10^-7 = x*x/0.1-x 1.47*10^-7 *(0.1-x) = x^2 x = 0.00012 [OH^-] = x = 0.00012M POH = -log[OH^-] = -log0.00012 = 3.92 PH = 14-POH = 14-3.92 = 10.08 .

What is the pH of a 0-1 M aqueous solution of NaH2PO4SolutionNaH2PO4 -.docx

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what is the ph of a 0.16M NaF(aq) solution? Solution NaF(aq) ------------> Na^+ (aq) + F^- (aq) 0.16M 0.16M F^- (aq) + H2O (l) ---------------> HF(aq) + OH^- (aq) I 0.16 0 0 C -x +x +x E 0.16-x +x +x Kb = Kw/Ka = 1*10^-14/6.8*10^-4 = 1.5*10^-11 Kb = [HF][OH^-]/[F^-] 1.5*10^-11 = x*x/0.16-x 1.5*10^-11 *(0.16-x) = x^2 x= 1.55*10^-6 [OH^-] = x = 1.55*10^-6M POH = -log[OH^-] = -log1.55*10^-6 = 5.81 PH = 14-POH = 14-5.81 = 8.19 >>>>answer .

what is the ph of a 0-16M NaF(aq) solution-SolutionNaF(aq) -----------.docx

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What is the pH of a 0.114 M solution of the acid HA if its ionization constant, K A , is 0.16 . Solution HA (aq) -------------> H^+ (aq) + A^- (aq) I 0.11 0 0 C -x +x +x E 0.11-x +x +x Ka = [H^+][A^-]/[HA] 0.16 = x*x/0.11-x 0.16*(0.11-x) = x^2 x = 0.075 [H^+] = x = 0.075M PH = -log[H^+] = -log0.075 = 1.125 >>>>>answer .

What is the pH of a 0-114 M solution of the acid HA if its ionization (1).docx

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What is the minimum number of nodes in an AVL tree of height 8? Let N(H) be the minimum number of nodes in an AVL tree of height h. What is the minimum number of nodes in an AVL tree of height 8? Let N(H) be the minimum number of nodes in an AVL tree of height h. Solution (* contains(t,x) is whether x is in the tree *) fun contains(t: tree, x: value): bool = case t of Empty => false | Node{value, left, right} => (case compare(x, value) of EQUAL => true | LESS => contains(left, x) | GREATER => contains(right, x)) (* add(t,x) is a BST with the same values as t, plus x *) fun add(t: tree, x: value): tree = let fun balance(t: tree): tree = t (* what to write here? *) in case t of Empty => Node{value=x, left=Empty, right=Empty} | Node {value, left, right} => (case compare(x, value) of EQUAL => Node{value=x, left=left, right=right} | LESS => Node{value=value, left=add(left, x), right=right} | GREATER => Node{value=value, left=left, right=add(right,x) } ) end .

What is the minimum number of nodes in an AVL tree of height 8- Let N(.docx

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