a concrete mix study crossing cement type with curing regime Given levels of factor A (a) = 5.0000; levels of factor B (b) = 4.0000; replicates per cell (n) = 3.0000, determine the total experimental runs (N) in runs.

Start with the thinking

• The governing relation printed in this handbook section is Runs in a two-factor factorial design.
• Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A two-factor factorial design tests every level combination of two factors with n replicates per cell.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$N = a b n$$

2. Step 2 — Rearrange symbolically for N:

   $$N = a b n$$

3. Step 3 — List the givens: levels of factor A (a) = 5.0000, levels of factor B (b) = 4.0000, replicates per cell (n) = 3.0000.

4. Step 4 — Substitute the given values:

   $$N = 5.0000 4.0000 3.0000$$

5. Step 5 — Evaluate:

   $$N = 60.0000\ \text{runs}$$

6. Step 6 — Check: returning N = 60.0000 runs to

   $$N = a b n$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 120.0 — kept a factor of two that cancels in the correct rearrangement.
• 30.0000 — dropped that same factor in the other direction.
• 66.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Two-factor Factorial Designs

an asphalt study crossing binder grade with compaction effort Given total experimental runs (N) = 56.0000 runs; levels of factor A (a) = 5.0000; levels of factor B (b) = 2.0000, determine the replicates per cell (n).

Start with the thinking

• The governing relation printed in this handbook section is Runs in a two-factor factorial design.
• Everything except n is given, so isolate n symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A two-factor factorial design tests every level combination of two factors with n replicates per cell.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$N = a b n$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \dfrac{N}{a b}$$

3. Step 3 — List the givens: total experimental runs (N) = 56.0000 runs, levels of factor A (a) = 5.0000, levels of factor B (b) = 2.0000.

4. Step 4 — Substitute the given values:

   $$n = \dfrac{56.0000}{5.0000 2.0000}$$

5. Step 5 — Evaluate:

   $$n = 5.6000$$

6. Step 6 — Check: returning n = 5.6000 to

   $$N = a b n$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 11.2000 — kept a factor of two that cancels in the correct rearrangement.
• 2.8000 — dropped that same factor in the other direction.
• 6.1600 — rounded an intermediate value before the final step.

Reference: FE Handbook — Two-factor Factorial Designs

a soil study crossing compaction energy with moisture level Given total experimental runs (N) = 140.0 runs; levels of factor A (a) = 4.0000; replicates per cell (n) = 5.0000, determine the levels of factor B (b).

Start with the thinking

• The governing relation printed in this handbook section is Runs in a two-factor factorial design.
• Everything except b is given, so isolate b symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A two-factor factorial design tests every level combination of two factors with n replicates per cell.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$N = a b n$$

2. Step 2 — Rearrange symbolically for b:

   $$b = \dfrac{N}{a n}$$

3. Step 3 — List the givens: total experimental runs (N) = 140.0 runs, levels of factor A (a) = 4.0000, replicates per cell (n) = 5.0000.

4. Step 4 — Substitute the given values:

   $$b = \dfrac{140.0}{4.0000 5.0000}$$

5. Step 5 — Evaluate:

   $$b = 7.0000$$

6. Step 6 — Check: returning b = 7.0000 to

   $$N = a b n$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 14.0000 — kept a factor of two that cancels in the correct rearrangement.
• 3.5000 — dropped that same factor in the other direction.
• 7.7000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Two-factor Factorial Designs

a concrete mix study crossing cement type with curing regime Given levels of factor A (a) = 5.0000; levels of factor B (b) = 4.0000; replicates per cell (n) = 2.0000, determine the total experimental runs (N) in runs.

Start with the thinking

• The governing relation printed in this handbook section is Runs in a two-factor factorial design.
• Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A two-factor factorial design tests every level combination of two factors with n replicates per cell.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$N = a b n$$

2. Step 2 — Rearrange symbolically for N:

   $$N = a b n$$

3. Step 3 — List the givens: levels of factor A (a) = 5.0000, levels of factor B (b) = 4.0000, replicates per cell (n) = 2.0000.

4. Step 4 — Substitute the given values:

   $$N = 5.0000 4.0000 2.0000$$

5. Step 5 — Evaluate:

   $$N = 40.0000\ \text{runs}$$

6. Step 6 — Check: returning N = 40.0000 runs to

   $$N = a b n$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 80.0000 — kept a factor of two that cancels in the correct rearrangement.
• 20.0000 — dropped that same factor in the other direction.
• 44.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Two-factor Factorial Designs

an asphalt study crossing binder grade with compaction effort Given total experimental runs (N) = 62.0000 runs; levels of factor A (a) = 4.0000; levels of factor B (b) = 2.0000, determine the replicates per cell (n).

Start with the thinking

• The governing relation printed in this handbook section is Runs in a two-factor factorial design.
• Everything except n is given, so isolate n symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A two-factor factorial design tests every level combination of two factors with n replicates per cell.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$N = a b n$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \dfrac{N}{a b}$$

3. Step 3 — List the givens: total experimental runs (N) = 62.0000 runs, levels of factor A (a) = 4.0000, levels of factor B (b) = 2.0000.

4. Step 4 — Substitute the given values:

   $$n = \dfrac{62.0000}{4.0000 2.0000}$$

5. Step 5 — Evaluate:

   $$n = 7.7500$$

6. Step 6 — Check: returning n = 7.7500 to

   $$N = a b n$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 15.5000 — kept a factor of two that cancels in the correct rearrangement.
• 3.8750 — dropped that same factor in the other direction.
• 8.5250 — rounded an intermediate value before the final step.

Reference: FE Handbook — Two-factor Factorial Designs

a soil study crossing compaction energy with moisture level Given total experimental runs (N) = 120.0 runs; levels of factor A (a) = 2.0000; replicates per cell (n) = 2.0000, determine the levels of factor B (b).

Start with the thinking

• The governing relation printed in this handbook section is Runs in a two-factor factorial design.
• Everything except b is given, so isolate b symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A two-factor factorial design tests every level combination of two factors with n replicates per cell.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$N = a b n$$

2. Step 2 — Rearrange symbolically for b:

   $$b = \dfrac{N}{a n}$$

3. Step 3 — List the givens: total experimental runs (N) = 120.0 runs, levels of factor A (a) = 2.0000, replicates per cell (n) = 2.0000.

4. Step 4 — Substitute the given values:

   $$b = \dfrac{120.0}{2.0000 2.0000}$$

5. Step 5 — Evaluate:

   $$b = 30.0000$$

6. Step 6 — Check: returning b = 30.0000 to

   $$N = a b n$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 60.0000 — kept a factor of two that cancels in the correct rearrangement.
• 15.0000 — dropped that same factor in the other direction.
• 33.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Two-factor Factorial Designs

a concrete mix study crossing cement type with curing regime Given levels of factor A (a) = 3.0000; levels of factor B (b) = 4.0000; replicates per cell (n) = 4.0000, determine the total experimental runs (N) in runs.

Start with the thinking

• The governing relation printed in this handbook section is Runs in a two-factor factorial design.
• Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A two-factor factorial design tests every level combination of two factors with n replicates per cell.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$N = a b n$$

2. Step 2 — Rearrange symbolically for N:

   $$N = a b n$$

3. Step 3 — List the givens: levels of factor A (a) = 3.0000, levels of factor B (b) = 4.0000, replicates per cell (n) = 4.0000.

4. Step 4 — Substitute the given values:

   $$N = 3.0000 4.0000 4.0000$$

5. Step 5 — Evaluate:

   $$N = 48.0000\ \text{runs}$$

6. Step 6 — Check: returning N = 48.0000 runs to

   $$N = a b n$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 96.0000 — kept a factor of two that cancels in the correct rearrangement.
• 24.0000 — dropped that same factor in the other direction.
• 52.8000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Two-factor Factorial Designs

an asphalt study crossing binder grade with compaction effort Given total experimental runs (N) = 178.0 runs; levels of factor A (a) = 3.0000; levels of factor B (b) = 3.0000, determine the replicates per cell (n).

Start with the thinking

• The governing relation printed in this handbook section is Runs in a two-factor factorial design.
• Everything except n is given, so isolate n symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A two-factor factorial design tests every level combination of two factors with n replicates per cell.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$N = a b n$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \dfrac{N}{a b}$$

3. Step 3 — List the givens: total experimental runs (N) = 178.0 runs, levels of factor A (a) = 3.0000, levels of factor B (b) = 3.0000.

4. Step 4 — Substitute the given values:

   $$n = \dfrac{178.0}{3.0000 3.0000}$$

5. Step 5 — Evaluate:

   $$n = 19.7778$$

6. Step 6 — Check: returning n = 19.7778 to

   $$N = a b n$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 39.5556 — kept a factor of two that cancels in the correct rearrangement.
• 9.8889 — dropped that same factor in the other direction.
• 21.7556 — rounded an intermediate value before the final step.

Reference: FE Handbook — Two-factor Factorial Designs

a soil study crossing compaction energy with moisture level Given total experimental runs (N) = 136.0 runs; levels of factor A (a) = 5.0000; replicates per cell (n) = 2.0000, determine the levels of factor B (b).

Start with the thinking

• The governing relation printed in this handbook section is Runs in a two-factor factorial design.
• Everything except b is given, so isolate b symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A two-factor factorial design tests every level combination of two factors with n replicates per cell.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$N = a b n$$

2. Step 2 — Rearrange symbolically for b:

   $$b = \dfrac{N}{a n}$$

3. Step 3 — List the givens: total experimental runs (N) = 136.0 runs, levels of factor A (a) = 5.0000, replicates per cell (n) = 2.0000.

4. Step 4 — Substitute the given values:

   $$b = \dfrac{136.0}{5.0000 2.0000}$$

5. Step 5 — Evaluate:

   $$b = 13.6000$$

6. Step 6 — Check: returning b = 13.6000 to

   $$N = a b n$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 27.2000 — kept a factor of two that cancels in the correct rearrangement.
• 6.8000 — dropped that same factor in the other direction.
• 14.9600 — rounded an intermediate value before the final step.

Reference: FE Handbook — Two-factor Factorial Designs

a concrete mix study crossing cement type with curing regime Given levels of factor A (a) = 4.0000; levels of factor B (b) = 5.0000; replicates per cell (n) = 5.0000, determine the total experimental runs (N) in runs.

Start with the thinking

• The governing relation printed in this handbook section is Runs in a two-factor factorial design.
• Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A two-factor factorial design tests every level combination of two factors with n replicates per cell.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$N = a b n$$

2. Step 2 — Rearrange symbolically for N:

   $$N = a b n$$

3. Step 3 — List the givens: levels of factor A (a) = 4.0000, levels of factor B (b) = 5.0000, replicates per cell (n) = 5.0000.

4. Step 4 — Substitute the given values:

   $$N = 4.0000 5.0000 5.0000$$

5. Step 5 — Evaluate:

   $$N = 100.0\ \text{runs}$$

6. Step 6 — Check: returning N = 100.0 runs to

   $$N = a b n$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 200.0 — kept a factor of two that cancels in the correct rearrangement.
• 50.0000 — dropped that same factor in the other direction.
• 110.0 — rounded an intermediate value before the final step.

Reference: FE Handbook — Two-factor Factorial Designs