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Question Number 99228 Answers: 1 Comments: 0

∫_0 ^∞ ((sin(x)ln(x))/x)dx=((−γπ)/2)

$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right){ln}\left({x}\right)}{{x}}{dx}=\frac{−\gamma\pi}{\mathrm{2}} \\ $$

Question Number 99227 Answers: 1 Comments: 2

Question Number 99225 Answers: 1 Comments: 0

A bag contains 4 tickets numbered 1, 2, 3, 4 and another bag contains 6 tickets numbered 2, 4, 6, 7, 8, 9. One bag is chosen and a ticket is drawn. The probability that the ticket bears the number 4 is

$$\mathrm{A}\:\mathrm{bag}\:\mathrm{contains}\:\mathrm{4}\:\mathrm{tickets}\:\mathrm{numbered}\:\mathrm{1},\:\mathrm{2}, \\ $$$$\mathrm{3},\:\mathrm{4}\:\mathrm{and}\:\mathrm{another}\:\mathrm{bag}\:\mathrm{contains}\:\mathrm{6}\:\mathrm{tickets} \\ $$$$\mathrm{numbered}\:\:\mathrm{2},\:\mathrm{4},\:\mathrm{6},\:\mathrm{7},\:\mathrm{8},\:\mathrm{9}.\:\mathrm{One}\:\mathrm{bag}\:\mathrm{is} \\ $$$$\mathrm{chosen}\:\mathrm{and}\:\mathrm{a}\:\mathrm{ticket}\:\mathrm{is}\:\mathrm{drawn}.\:\mathrm{The} \\ $$$$\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{ticket}\:\mathrm{bears}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{4}\:\mathrm{is} \\ $$

Question Number 99222 Answers: 0 Comments: 0

calculate Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 (n+2)^3 ))

$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{3}} } \\ $$

Question Number 99218 Answers: 0 Comments: 3

Question Number 99205 Answers: 2 Comments: 1

Question Number 99194 Answers: 1 Comments: 0

Question Number 99193 Answers: 1 Comments: 4

Question Number 99175 Answers: 2 Comments: 0

A man and a woman have 3 boys and 3 girls. (i) in how many ways will they sit in a row such that the 3 boys and 3 girls are inbetween the man and woman. (ii) Say the man decides of the 3 boys and 3 girls he has 2 of the kids should help him out in a project, in how many ways can this be done, if heyoungest boy and oldest girl can′t join.

$$\:\:\mathrm{A}\:\mathrm{man}\:\mathrm{and}\:\mathrm{a}\:\mathrm{woman}\:\mathrm{have}\:\mathrm{3}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{3}\:\mathrm{girls}.\: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{in}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{will}\:\mathrm{they}\:\mathrm{sit}\:\mathrm{in}\:\mathrm{a}\:\mathrm{row}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{the}\:\mathrm{3}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{3}\:\mathrm{girls}\:\mathrm{are}\:\mathrm{inbetween}\:\mathrm{the}\:\mathrm{man}\:\mathrm{and}\:\mathrm{woman}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Say}\:\mathrm{the}\:\mathrm{man}\:\mathrm{decides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{3}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{3}\:\mathrm{girls}\:\mathrm{he}\:\mathrm{has}\:\mathrm{2}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{kids}\:\mathrm{should}\:\mathrm{help}\:\mathrm{him}\:\mathrm{out}\:\mathrm{in}\:\mathrm{a}\:\mathrm{project}, \\ $$$$\mathrm{in}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{done},\:\mathrm{if}\:\mathrm{heyoungest}\:\mathrm{boy}\:\mathrm{and}\:\mathrm{oldest} \\ $$$$\mathrm{girl}\:\mathrm{can}'\mathrm{t}\:\mathrm{join}. \\ $$

Question Number 99171 Answers: 0 Comments: 3

Question Number 99168 Answers: 1 Comments: 0

Question Number 99159 Answers: 0 Comments: 2

Question Number 99154 Answers: 4 Comments: 0

Question Number 99146 Answers: 1 Comments: 0

1) explicit f(a) =∫_1 ^(√3) arctan((a/x))dx with a>0 2) calculate ∫_1 ^(√3) arctan((2/x))dx and ∫_1 ^(√3) arctan((3/x))dx

$$\left.\mathrm{1}\right)\:\mathrm{explicit}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:\mathrm{arctan}\left(\frac{\mathrm{a}}{\mathrm{x}}\right)\mathrm{dx}\:\:\mathrm{with}\:\mathrm{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{calculate}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\mathrm{arctan}\left(\frac{\mathrm{2}}{\mathrm{x}}\right)\mathrm{dx}\:\mathrm{and}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right)\mathrm{dx} \\ $$

Question Number 99142 Answers: 1 Comments: 0

If 4 dice are thrown together, then the probability that the sum of the numbers appearing on them is 13, is

$$\mathrm{If}\:\mathrm{4}\:\mathrm{dice}\:\mathrm{are}\:\mathrm{thrown}\:\mathrm{together},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers} \\ $$$$\mathrm{appearing}\:\mathrm{on}\:\mathrm{them}\:\mathrm{is}\:\mathrm{13},\:\mathrm{is} \\ $$

Question Number 99141 Answers: 1 Comments: 0

The probability that in a random arrangement of the letters of the word ′UNIVERSITY′ the two I′ do not come together is

$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{in}\:\mathrm{a}\:\mathrm{random} \\ $$$$\mathrm{arrangement}\:\mathrm{of}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word} \\ $$$$'\mathrm{UNIVERSITY}'\:\mathrm{the}\:\mathrm{two}\:\mathrm{I}'\:\mathrm{do}\:\mathrm{not}\:\mathrm{come} \\ $$$$\mathrm{together}\:\mathrm{is} \\ $$

Question Number 99139 Answers: 1 Comments: 0

One ticket is selected at random from 100 tickets numbered 00, 01, 02, ..., 99 Suppose A and B are the sum and product of the digit found on the ticket. Then P (A = 7/B= 0) is given by

$$\mathrm{One}\:\mathrm{ticket}\:\mathrm{is}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random}\:\mathrm{from} \\ $$$$\mathrm{100}\:\mathrm{tickets}\:\mathrm{numbered}\:\mathrm{00},\:\mathrm{01},\:\mathrm{02},\:...,\:\mathrm{99} \\ $$$$\mathrm{Suppose}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{and}\: \\ $$$$\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digit}\:\mathrm{found}\:\mathrm{on}\:\mathrm{the}\:\mathrm{ticket}. \\ $$$$\mathrm{Then}\:{P}\:\left({A}\:=\:\mathrm{7}/{B}=\:\mathrm{0}\right)\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$

Question Number 99138 Answers: 1 Comments: 0

If A and B are independent events and P(C)=0, then

$$\mathrm{If}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are}\:\mathrm{independent}\:\mathrm{events}\:\mathrm{and} \\ $$$${P}\left({C}\right)=\mathrm{0},\:\mathrm{then} \\ $$

Question Number 99137 Answers: 0 Comments: 0

If m rupee coins and n ten paise coins are placed in a line, then the probability that the extreme coins are ten paise coins is

$$\mathrm{If}\:{m}\:\mathrm{rupee}\:\mathrm{coins}\:\mathrm{and}\:{n}\:\mathrm{ten}\:\mathrm{paise}\:\mathrm{coins} \\ $$$$\mathrm{are}\:\mathrm{placed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{line},\:\mathrm{then}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{extreme}\:\mathrm{coins}\:\mathrm{are}\:\mathrm{ten}\:\mathrm{paise}\:\mathrm{coins} \\ $$$$\mathrm{is} \\ $$

Question Number 99136 Answers: 0 Comments: 0

Three persons A, B, C are to speak at a function along with five others. If they all speak in random order, the probqbility that A speaks before B and B speaks before C is

$$\mathrm{Three}\:\mathrm{persons}\:{A},\:{B},\:{C}\:\:\mathrm{are}\:\mathrm{to}\:\mathrm{speak}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{along}\:\mathrm{with}\:\mathrm{five}\:\mathrm{others}.\:\mathrm{If}\:\mathrm{they} \\ $$$$\mathrm{all}\:\mathrm{speak}\:\mathrm{in}\:\mathrm{random}\:\mathrm{order},\:\mathrm{the}\:\mathrm{probqbility} \\ $$$$\mathrm{that}\:{A}\:\mathrm{speaks}\:\mathrm{before}\:{B}\:\mathrm{and}\:{B}\:\mathrm{speaks}\: \\ $$$$\mathrm{before}\:{C}\:\mathrm{is} \\ $$

Question Number 99133 Answers: 0 Comments: 0

Equilibrate it using oxydation′s number (NH_4 )_2 Cr_2 O_7 →N_2 +H_2 O+Cr_(2 ) O_7

$${Equilibrate}\:{it}\:{using}\:{oxydation}'{s} \\ $$$${number} \\ $$$$\left({NH}_{\mathrm{4}} \right)_{\mathrm{2}} {Cr}_{\mathrm{2}} {O}_{\mathrm{7}} \rightarrow{N}_{\mathrm{2}} +{H}_{\mathrm{2}} {O}+{Cr}_{\mathrm{2}\:\:} {O}_{\mathrm{7}} \\ $$

Question Number 99123 Answers: 2 Comments: 1

Question Number 99120 Answers: 0 Comments: 2

prove that: ∫_(−(1/2)) ^∞ e^(−(4x^6 +12x^5 +15x^4 +10x^3 +4x^2 +x)) dx =((e)^(1/8) /3)[((Γ((1/6))^((−1)/2) )/(2(2)^(1/3) ))1F2(_(1/3,2/3) ^(1/6) ∣((−1)/(69/2)) ) +((Γ(5/6))/(128(4)^(1/3) ))1F2(_(4/3,5/3) ^(5/6) ∣((−1)/(69/2))) −((√π)/(16))12(_(2/3,4/3) ^(1/2) ∣((−1)/(69/2)))

$${prove}\:{that}: \\ $$$$\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\infty} {e}^{−\left(\mathrm{4}{x}^{\mathrm{6}} +\mathrm{12}{x}^{\mathrm{5}} +\mathrm{15}{x}^{\mathrm{4}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +{x}\right)} {dx} \\ $$$$=\frac{\sqrt[{\mathrm{8}}]{{e}}}{\mathrm{3}}\left[\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}}\mathrm{1}{F}\mathrm{2}\left(_{\mathrm{1}/\mathrm{3},\mathrm{2}/\mathrm{3}} ^{\mathrm{1}/\mathrm{6}} \mid\frac{−\mathrm{1}}{\mathrm{69}/\mathrm{2}}\:\right)\:+\frac{\Gamma\left(\mathrm{5}/\mathrm{6}\right)}{\mathrm{128}\sqrt[{\mathrm{3}}]{\mathrm{4}}}\mathrm{1}{F}\mathrm{2}\left(_{\mathrm{4}/\mathrm{3},\mathrm{5}/\mathrm{3}} ^{\mathrm{5}/\mathrm{6}} \mid\frac{−\mathrm{1}}{\mathrm{69}/\mathrm{2}}\right)\:−\frac{\sqrt{\pi}}{\mathrm{16}}\mathrm{12}\left(_{\mathrm{2}/\mathrm{3},\mathrm{4}/\mathrm{3}} ^{\mathrm{1}/\mathrm{2}} \mid\frac{−\mathrm{1}}{\mathrm{69}/\mathrm{2}}\right)\:\right. \\ $$

Question Number 99118 Answers: 4 Comments: 0

Question Number 99117 Answers: 1 Comments: 0

let a,b,c ∈R determine the minimum value ((3a)/(b+c))+((4b)/(a+c))+((5c)/(a+b))

$${let}\:{a},{b},{c}\:\in\mathbb{R}\:{determine}\:{the}\:{minimum} \\ $$$${value} \\ $$$$ \\ $$$$\frac{\mathrm{3}{a}}{{b}+{c}}+\frac{\mathrm{4}{b}}{{a}+{c}}+\frac{\mathrm{5}{c}}{{a}+{b}} \\ $$

Question Number 99114 Answers: 1 Comments: 0

calculate: ∫(√x)sinh^(−1) (x)dx where sinh^(−1) (x) is the inverse hyperbolic sine function

$${calculate}: \\ $$$$\int\sqrt{{x}}{sinh}^{−\mathrm{1}} \left({x}\right){dx} \\ $$$${where}\:{sinh}^{−\mathrm{1}} \left({x}\right)\:{is}\:{the}\:{inverse}\:{hyperbolic}\: \\ $$$${sine}\:{function} \\ $$$$ \\ $$$$ \\ $$

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