A popular candy is manufactured to weigh 0.16 ounces per piece. The pieces aresold in packages of 36. If the allowable amount of variation in weight for onepiece of candy is 0.03 ounces, which of the following equations can be used todetermine the highest and lowest allowable weight, w, of a package of 36 pieces?Assume that the weight of the packaging is negligible a. â£wâ0.16â£/36=0.03
b. â£wâ5.76â£/36=0.03
c. 36 |wâ0.16| =0.03
d. 36|w-5.76|=0.03

B.) |w - 5.76| / 36 = 0.03

Step-by-step explanation:

Given :

Weight of individual candy piece = 0.16 ounce

Number of candy pieces per package = 36

Allowable variation in weight of individual piece = 0.03 ounces

The highest and lowest allowable weight, w of a package of 36 pieces ;

Actual manufacturing weight of entire 36 candy = (36 * 0.16) = 5.76

Allowable weight ± actual weight ≤ allowable variation

w ± 0.16 ≤ 0.03

|w - 0.16| ≤ 0.03

For the entire 36 candy pieces :

|w - 0.16*36| = 0.03 * 36

|w - 5.76| = 1.08

|w - 0.16*36| = 0.03 * 36

Could also be written as :

|w - 5.76| / 36 = 0.03

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841