Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

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 :

jcherry99

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.

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.