how much is 47 grams of 14 karat gold?

how much is 47 grams of 14 karat gold?
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1 Answer

Answer :

it is worth approxiamately 921.75
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Related questions

What is the difference between 12 karat and 14 karat gold?

Description : What is the difference between 12 karat and 14 karat gold?

Last Answer : 14 Karat gold contains more pure gold in it than 12 karat, and is therefore worth more money, however 12 karat gold is harder, so it will be more durable and less easy to scratch or damage.

What is the difference between 10 and 14 karat gold used in a ring?

Description : What is the difference between 10 and 14 karat gold used in a ring?

Last Answer : 10 karat gold is 10 parts of pure gold to 14 parts of other metal and 14 karat gold is 14 parts of pure gold.

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Medium 1 is a glass (n1 = 1.5) and medium 2 is an ethyl alcohol (n2 = 1.36). For an angle of incidence of 30 degrees, determine the angle of refraction. ∙ A. 44.5 degrees ∙ B. 14.56 degrees ∙ C. 33.47 degrees ∙ D. 75 degrees

Description : Medium 1 is a glass (n1 = 1.5) and medium 2 is an ethyl alcohol (n2 = 1.36). For an angle of incidence of 30 degrees, determine the angle of refraction. ∙ A. 44.5 degrees ∙ B. 14.56 degrees ∙ C. 33.47 degrees ∙ D. 75 degrees

Last Answer : C. 33.47 degrees

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Is there Zakat on the money of the association ? I owe Rs 50,000 / - to one and Rs 47,000 / - to another. We have an association where I have deposited Rs. 150,000 / -. The money of the association is deposited in the bank , no work is done. Do I have to pay zakat ? How much more do you have to pay ?

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How much will the length of a 1.0 km section of concrete highway change if the temperature varies from -15˚C in winter to 41˚C in summer?  a. 0.67 m  b. 2.2 m  c. 3.1 m  d. 0.47 m

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My mother is three times as old as I am. Together, we are 47 years old. How old is my mother?

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Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Description : Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

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If P (event E) = 0.47, then find P(not E). -Maths 9th

Description : If P (event E) = 0.47, then find P(not E). -Maths 9th

Last Answer : P(not E) = 1 - P(E) ⇒ 1 - 0.47 = 0.53

Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Description : Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Last Answer : Mean of 36 observations = 12 Total of 36 observations = 36 x 12 = 432 Correct sum of 36 observations = 432 – 74 + 47 = 405 Correct mean of 36 observations = 405/ 36 =11.25

If P (event E) = 0.47, then find P(not E). -Maths 9th

Description : If P (event E) = 0.47, then find P(not E). -Maths 9th

Last Answer : P(not E) = 1 - P(E) ⇒ 1 - 0.47 = 0.53

Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Description : Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Last Answer : Let x = 0.7Bar ⇒ x = 0.777.......... .........(1) Multiplying (1) by 10 ⇒ 10x = 7.7...... = 7.777 ...........(2) Subtracting (1) from (2) ⇒ 10x - x = 9x ⇒ 7.777 - 0.777 = 7 ... and then solving it, we get. ⇒ (594 + 770 + 470)/990 ⇒ 1834/990 ⇒ 917/495 Hence, 0.6 + 0.7Bar + 0.47Bar = 917/495

Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Description : Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Last Answer : Let x=0.666....... (1) Multiply equation (1 by 10 10x = 6.666....... (2) Subtract equation (1) from (2) x=6/9 Similarly 0.7bar =7/9 and 0.47bar = 47/99. 6/9+7/9+47/99=190/99

Teacher asked the students 'Can we write 0.47(recurring) -Maths 9th

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Convert each of the following fraction into a percentage: (i) 47 100 (ii) 9 20 (iii) 3 8 (iv) 8 125 (v) 19 500 (vi) 4 15 (vii) 2 3 (viii) 1 3 5 -Maths

Description : Convert each of the following fraction into a percentage: (i) 47 100 (ii) 9 20 (iii) 3 8 (iv) 8 125 (v) 19 500 (vi) 4 15 (vii) 2 3 (viii) 1 3 5 -Maths

Last Answer : We have the following: (i) 47100=47100×100% = 47% (ii) 920=920×100% =9×5%= 45% (iii) 38=38×100% =3×252%=752%=3712% (iv) 8125=8125×100%=8×45%=325%=6.4% (v) ... ;203%=803%=2623% (vii) 23=23×100%=2003%=6623% (viii) 135=85=85×100%=(8×20)%=160%

IfP (event E) = 0.47, then find P(not E). -Maths 9th

Description : IfP (event E) = 0.47, then find P(not E). -Maths 9th

Last Answer : P(not E) = 1 - P(E) ⇒ 1 - 0.47 = 0.53

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How can 11-04-47, 11-11-72, 09-23-75 and 09-16-79 all fall on the same day of the week?

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Description : XI. Answer any TWO of the following questions.47) On the basis of data you have collected regarding participation of women in your locality in below mentioned catagories construct a pie chart.a) Self employed.b) Regular salaried employees.c) Causel wage labours.

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Description : The mode of the data 48, 45, 49, 51, 46, 48, 47, 49, 48, and 51 is ______.

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Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \frac{3}{4} \)b) \( \frac{5}{8} \)c) \( \frac{9}{8} \)d) \( \frac{7}{8} \)47. The sum of perimeter of all triangle is (in \( cm \) )a) 144b) 625c) 400d) 16948. The area of all the triangle is (in sq cm)a) 576b) \( 144 \sqrt{3} \)c) \( 169 \sqrt{3} \)d) \( 192 \sqrt{3} \)49. The sum of perimeter of first 6 triangle is (in \( cm \) )a) 120b) \( \frac{567}{4} \)c) \( \frac{569}{4} \)d) 14450. The side of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3

Description : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \ ... of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3

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A single electron orbits around a stationary nucleus of charge `+Ze`, where Z is a constant and e is the electronic charge. It requires `47.2 eV` to e

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