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If you have 58 grams of substance A and 19 grams of substance B and 19 grams of substance C and 15 grams of substance D and 10 grams of substance E and you mix all of these powders perfectly in a bag and you then take a 4 gram serving size out of the already mixed bag how much of each substance would you have in milligrams of each substance from the 4 gram serving size?

Sagot :

Jcherry99idn

Answer:

Step-by-step explanation:

Givens

58 grams of A

19 grams of B

19 grams of C

15 grams of D

10 grams of E

Total 58 + 19 + 19 + 15 + 10 = 121

Solution

58/121 = A/4              Cross multiply

4 * 58 = 121A             Combine

232 = 121 A                Divide by 121

232/121 = 121x/121      

A = 1.92

The others are done with same way

19/121 = x / 4

19*4 = 121 B

76 = 121 B

76/121 = 121/121 B

B = .628

C is the same as be

C = 0.628

15/121 = C / 4

60 = 121 * C

60/121 = D

D = .496

10/121 = E / 4

40 * 4 = 121 * E

40 = 121 * E

E = 40/121

E = .331

Total

A + B + C + D + E

Total = l.92+0.628 + 0.628 + 0.496 + 0.331

Total = 4.003 which is close enough to 4.

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