Q = d / d+ n on a manufacturer's assembly line, d parts are found to be defective and n parts are nondefective. the formula above is used to calculate a quality-of-parts ratio. What is d expressed in terms of the other two variables? a. n / 1 - q
b. nq / 1 - q
c. n / q-1
d. nq / q-1

d. nq / q - 1

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra I

Factoring

Step-by-step explanation:

Step 1: Define

q = d / (d + n)

Step 2: Solve for d

Multiply (d + n) on both sides:                    q(d + n) = dDistribute q:                                                 dq + qn = dSubtract qn on both sides:                         dq = d - qnSubtract d on both sides:                           dq - d = -qnFactor GCF:                                                 d(q - 1) = -qnDivide (q - 1) on both sides:                        d = -qn / (q - 1)Rearrange:                                                   d = nq / -(q - 1)Distribute -1:                                                 d = nq / 1 - q

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< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)