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Question Number 170810    Answers: 0   Comments: 0

The mean height of a population of girls aged 15 to 19 years in a northern province in Ghana was found to be 165 cm with a standard deviation of 15 cm. Assuming that the heights are normally distributed, find the heights in centimetres that correspond to the following percentiles: a. Between the 20th and 50th percentiles.

$$ \\ $$The mean height of a population of girls aged 15 to 19 years in a northern province in Ghana was found to be 165 cm with a standard deviation of 15 cm. Assuming that the heights are normally distributed, find the heights in centimetres that correspond to the following percentiles: a. Between the 20th and 50th percentiles.

Question Number 170809    Answers: 2   Comments: 0

Question Number 170802    Answers: 2   Comments: 0

Question Number 170800    Answers: 1   Comments: 0

Prove that, for any real number x and odd positive integer n, cos^n x=(1/2^(n+1) )Σ_(k=0) ^((n−1)/2) C_k ^n cos (n−2k)x

$$\mathrm{Prove}\:\mathrm{that},\:\mathrm{for}\:\mathrm{any}\:\mathrm{real}\:\mathrm{number}\:{x}\:\mathrm{and}\:\mathrm{odd}\:\mathrm{positive}\:\mathrm{integer}\:{n}, \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}^{{n}} {x}=\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{cos}\:\left({n}−\mathrm{2}{k}\right){x} \\ $$

Question Number 170796    Answers: 2   Comments: 0

Question Number 170795    Answers: 2   Comments: 0

solve (x^2 +y^2 +1)dx+x(x−2y)dy=0

$${solve} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right){dx}+{x}\left({x}−\mathrm{2}{y}\right){dy}=\mathrm{0} \\ $$

Question Number 170794    Answers: 2   Comments: 0

2n objects of each of three kinds are given to two persons, so that each person gets 3n objects. Prove that this can be done in 3n^2 + 3n + 1 ways.

$$\mathrm{2}{n}\:\mathrm{objects}\:\mathrm{of}\:\mathrm{each}\:\mathrm{of}\:\mathrm{three}\:\mathrm{kinds}\:\mathrm{are} \\ $$$$\mathrm{given}\:\mathrm{to}\:\mathrm{two}\:\mathrm{persons},\:\mathrm{so}\:\mathrm{that}\:\mathrm{each} \\ $$$$\mathrm{person}\:\mathrm{gets}\:\mathrm{3}{n}\:\mathrm{objects}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{in}\:\mathrm{3}{n}^{\mathrm{2}} \:+\:\mathrm{3}{n}\:+\:\mathrm{1}\:\mathrm{ways}. \\ $$

Question Number 170781    Answers: 2   Comments: 0

(√((2−x))) = (2−x)^2 solve for x

$$\sqrt{\left(\mathrm{2}−\mathrm{x}\right)}\:\:\:=\:\:\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} \:\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\:\:\:\:\:\boldsymbol{\mathrm{x}} \\ $$

Question Number 170780    Answers: 1   Comments: 0

Question Number 170776    Answers: 0   Comments: 2

Question Number 170772    Answers: 0   Comments: 0

Question Number 170767    Answers: 2   Comments: 0

Question Number 170766    Answers: 2   Comments: 0

lim_(x→−∞) ((x^7 +x^6 −1))^(1/7) +((x^5 +1−x^9 ))^(1/9) =?

$$\:\:\:\:\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{7}}]{{x}^{\mathrm{7}} +{x}^{\mathrm{6}} −\mathrm{1}}\:+\sqrt[{\mathrm{9}}]{{x}^{\mathrm{5}} +\mathrm{1}−{x}^{\mathrm{9}} }\:=? \\ $$

Question Number 170754    Answers: 1   Comments: 5

Solve: x + (√y) = 3 ...... (i) (√x) + y = 5 ...... (ii)

$$\mathrm{Solve}: \\ $$$$\:\:\:\:\:\mathrm{x}\:\:\:\:+\:\:\:\sqrt{\mathrm{y}}\:\:\:\:=\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:......\:\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\sqrt{\mathrm{x}}\:\:\:\:+\:\:\:\:\mathrm{y}\:\:\:\:=\:\:\:\:\mathrm{5}\:\:\:\:\:\:\:\:......\:\:\left(\mathrm{ii}\right) \\ $$

Question Number 170753    Answers: 0   Comments: 0

Question Number 170752    Answers: 2   Comments: 0

Solve:− ((sinx + cosx)/(sinx − cosx)) = (((√(3 )) − 1)/( (√(3 )) + 1))

$${Solve}:−\:\:\frac{{sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}}{{sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}}\:=\:\frac{\sqrt{\mathrm{3}\:}\:−\:\mathrm{1}}{\:\sqrt{\mathrm{3}\:}\:+\:\mathrm{1}}\: \\ $$

Question Number 170748    Answers: 1   Comments: 0

A number of four different digits is to be formed from the digits 1,2,3,4,5,6,7,8,9.Find how many of them are: i)greater than 4000 ii)less than 6000 iii) divisible by 2 iv) divisible by 5

$$ \\ $$A number of four different digits is to be formed from the digits 1,2,3,4,5,6,7,8,9.Find how many of them are: i)greater than 4000 ii)less than 6000 iii) divisible by 2 iv) divisible by 5

Question Number 170743    Answers: 0   Comments: 0

2. Uma soluca^ o tampao foi preparada misturarando 200ml NH_(3 ) 0,6 moles e 300ml de uma solucao de NH_3 Cl 0,2 moles P^(kh) =9,24 e log 2=0,3 a) Qual e o P^h desta solucao tampa^ o supondo-se um volume de 500ml? b)Qual sera o P^h depois de ser adicionado 0,2 molar de ion [H^− ] Dados n(CH_3 -CH_2 COOH)=0,02mok/l=0,02M Ka=1,3∙10^(−5) n(CH_3 −CH_2 -COONa)=0,015mol/l=0,015M V=1l log 2=0,3 log 0,11 P^h =? 1°Passo 2°Passo P^(ka) =−log Ka P^h =P^(ka) +log(([Base])/([Acido])) P^(ka) =−log 1,3∙10^(−5) ? P^h =4,89+log(([0,02])/([0,015])) P^(ka) =(−5+0,11) P^h =4,89+log1,3 P^(ka) =5−11 P^h =4,89+0,11 P^(ka) =4,49 P^h =5 3°Passo P^h =P^h +log(([Basica])/([Acida])) P^h =4,89+log(([0,02])/([0,025])) P^h =4,89+log0,8 P^h =4,89+0,096 P^h =4,986≈5

$$ \\ $$$$\mathrm{2}.\:\mathrm{Uma}\:\:\mathrm{soluc}\overset{ } {\mathrm{a}o}\:\:\mathrm{tampao}\:\mathrm{foi}\:\mathrm{preparada}\:\mathrm{misturarando}\:\mathrm{200}\boldsymbol{\mathrm{m}{l}}\:\mathrm{NH}_{\mathrm{3}\:} \:\:\mathrm{0},\mathrm{6}\:{moles} \\ $$$${e}\:\mathrm{300}{m}\boldsymbol{{l}}\:{de}\:{uma}\:{solucao}\:{de}\:\mathrm{NH}_{\mathrm{3}} \mathrm{C}{l}\:\:\:\mathrm{0},\mathrm{2}\:{moles}\:{P}^{\mathrm{kh}} =\mathrm{9},\mathrm{24}\:{e}\:\mathrm{log}\:\mathrm{2}=\mathrm{0},\mathrm{3} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Q}{ual}\:{e}\:{o}\:{P}^{\mathrm{h}} \:\mathrm{desta}\:\mathrm{solucao}\:\mathrm{tamp}\overset{ } {\mathrm{a}o}\:\mathrm{supondo}-{se}\:{um}\:{volume}\:{de}\:\mathrm{500}\boldsymbol{{ml}}? \\ $$$$\left.\mathrm{b}\right)\mathrm{Q}{ual}\:{sera}\:{o}\:{P}^{\mathrm{h}} \:{depois}\:{de}\:{ser}\:{adicionado}\:\mathrm{0},\mathrm{2}\:{molar}\:{de}\:{ion}\:\left[\mathrm{H}^{−} \right] \\ $$$$\mathrm{Dados} \\ $$$$\mathrm{n}\left(\mathrm{CH}_{\mathrm{3}} -\mathrm{CH}_{\mathrm{2}} \mathrm{COOH}\right)=\mathrm{0},\mathrm{02}\boldsymbol{{mok}}/\boldsymbol{{l}}=\mathrm{0},\mathrm{02M} \\ $$$${K}\mathrm{a}=\mathrm{1},\mathrm{3}\centerdot\mathrm{10}^{−\mathrm{5}} \\ $$$${n}\left(\mathrm{CH}_{\mathrm{3}} −\mathrm{CH}_{\mathrm{2}} -\mathrm{COONa}\right)=\mathrm{0},\mathrm{015}{mol}/\boldsymbol{{l}}=\mathrm{0},\mathrm{015M} \\ $$$$\mathrm{V}=\mathrm{1}\boldsymbol{{l}} \\ $$$$\mathrm{log}\:\mathrm{2}=\mathrm{0},\mathrm{3} \\ $$$$\mathrm{log}\:\mathrm{0},\mathrm{11} \\ $$$$\mathrm{P}^{\mathrm{h}} =? \\ $$$$\:\:\mathrm{1}°\boldsymbol{\mathrm{Passo}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}°\mathrm{Passo} \\ $$$$\mathrm{P}^{\mathrm{ka}} =−\mathrm{log}\:\mathrm{Ka}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{P}^{\mathrm{ka}} +\mathrm{log}\frac{\left[\mathrm{Base}\right]}{\left[\mathrm{Acido}\right]} \\ $$$$\mathrm{P}^{\mathrm{ka}} =−\mathrm{log}\:\mathrm{1},\mathrm{3}\centerdot\mathrm{10}^{−\mathrm{5}} \:\:\:\:\:?\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log}\frac{\left[\mathrm{0},\mathrm{02}\right]}{\left[\mathrm{0},\mathrm{015}\right]} \\ $$$$\mathrm{P}^{\mathrm{ka}} =\left(−\mathrm{5}+\mathrm{0},\mathrm{11}\right)\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log1},\mathrm{3} \\ $$$$\mathrm{P}^{\mathrm{ka}} =\mathrm{5}−\mathrm{11}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{0},\mathrm{11} \\ $$$$\mathrm{P}^{\mathrm{ka}} =\mathrm{4},\mathrm{49}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{5} \\ $$$$\mathrm{3}°\boldsymbol{\mathrm{Passo}} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{P}^{\mathrm{h}} +\mathrm{log}\frac{\left[\mathrm{Basica}\right]}{\left[\mathrm{Acida}\right]} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log}\frac{\left[\mathrm{0},\mathrm{02}\right]}{\left[\mathrm{0},\mathrm{025}\right]} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log0},\mathrm{8} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{0},\mathrm{096} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{986}\approx\mathrm{5} \\ $$

Question Number 170738    Answers: 0   Comments: 9

Question Number 170737    Answers: 1   Comments: 0

(3/(sin^2 40°)) − (1/(cos^2 40°)) + 64 sin^2 40° = ?

$$\frac{\mathrm{3}}{\mathrm{sin}^{\mathrm{2}} \mathrm{40}°}\:−\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{40}°}\:+\:\mathrm{64}\:\mathrm{sin}^{\mathrm{2}} \mathrm{40}°\:\:=\:\:? \\ $$

Question Number 170729    Answers: 3   Comments: 2

Question Number 170726    Answers: 0   Comments: 0

Question Number 170725    Answers: 1   Comments: 0

Question Number 170722    Answers: 2   Comments: 0

Question Number 170721    Answers: 0   Comments: 1

Which is the true magnefication′s formula for the concave mirror? m=(f/(p−f))? m=−(f/(p−f))?

$${Which}\:{is}\:{the}\:{true}\:{magnefication}'{s}\: \\ $$$${formula}\:{for}\:{the}\:{concave}\:{mirror}? \\ $$$${m}=\frac{{f}}{{p}−{f}}? \\ $$$${m}=−\frac{{f}}{{p}−{f}}? \\ $$

Question Number 170720    Answers: 0   Comments: 0

Prove the magnefication for concave mirror. m=(f/(∣f−p∣))

$${Prove}\:{the}\:{magnefication}\:{for}\:{concave} \\ $$$${mirror}. \\ $$$${m}=\frac{{f}}{\mid{f}−{p}\mid} \\ $$

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