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SUBJECT 1 — COLLEGE ALGEBRA

Question 1: Solving a Quadratic Equation

Solve for x: 2x² − 7x + 3 = 0

Step 1. Identify the coefficients for the quadratic formula.

a = 2, b = −7, c = 3

Step 2. Compute the discriminant, b² − 4ac.

b² − 4ac = (−7)² − 4(2)(3) = 49 − 24 = 25

Because the discriminant is positive and a perfect square, there are two distinct rational roots.

Step 3. Apply the quadratic formula x = [−b ± √(b² − 4ac)] / (2a).

x = [7 ± √25] / (2·2) = [7 ± 5] / 4

Step 4. Evaluate both roots.

x = (7 + 5)/4 = 12/4 = 3

x = (7 − 5)/4 = 2/4 = 1/2

Answer: x = 3 or x = 1/2

SUBJECT 2 — INTRODUCTORY STATISTICS

Question 2: Confidence Interval for a Population Mean

A sample of n = 36 light bulbs has a mean lifetime of 1,200 hours. The population standard deviation is known to be σ = 90 hours. Construct a 95% confidence interval for the true mean lifetime.

Step 1. Identify the known values.

x̄ = 1200, σ = 90, n = 36, confidence = 95%

Step 2. Find the critical z-value. For 95% confidence, the central area is 0.95, leaving 0.025 in each tail.

z* = 1.96

Step 3. Compute the standard error of the mean, σ / √n.

SE = 90 / √36 = 90 / 6 = 15

Step 4. Compute the margin of error, z* × SE.

E = 1.96 × 15 = 29.4

Step 5. Build the interval x̄ ± E.

1200 − 29.4 = 1170.6

1200 + 29.4 = 1229.4

Answer: The 95% confidence interval is (1170.6, 1229.4) hours.

Interpretation: we are 95% confident that the true mean lifetime of all bulbs lies between about 1,170.6 and 1,229.4 hours.

SUBJECT 3 — GENERAL CHEMISTRY

Question 3: Preparing a Diluted Solution

How many milliliters of a 12.0 M stock HCl solution are needed to prepare 500.0 mL of a 0.150 M HCl solution?

Step 1. Recognize this as a dilution problem and write the dilution equation.

M₁V₁ = M₂V₂

Step 2. List the known quantities (the stock is solution 1, the diluted target is solution 2).

M₁ = 12.0 M (concentrated stock)

M₂ = 0.150 M, V₂ = 500.0 mL (final solution)

Step 3. Solve the equation for V₁.

V₁ = (M₂V₂) / M₁

V₁ = (0.150 M × 500.0 mL) / 12.0 M

V₁ = 75.0 / 12.0 = 6.25 mL

Answer: Measure 6.25 mL of the 12.0 M stock, then add water up to a total of 500.0 mL.

Safety note: always add acid to water, never water to acid, to dissipate heat safely.

SUBJECT 4 — PRINCIPLES OF MICROECONOMICS

Question 4: Price Elasticity of Demand

When the price of a product rises from $20 to $24, quantity demanded falls from 1,000 to 800 units. Use the midpoint method to find the price elasticity of demand and classify it.

Step 1. Recall the midpoint (arc) elasticity formula, which uses averages so the result is the same in either direction.

E = (ΔQ / avg Q) ÷ (ΔP / avg P)

Step 2. Compute the change and average for quantity.

ΔQ = 800 − 1000 = −200

avg Q = (1000 + 800) / 2 = 900

%ΔQ = −200 / 900 = −0.2222

Step 3. Compute the change and average for price.

ΔP = 24 − 20 = 4

avg P = (20 + 24) / 2 = 22

%ΔP = 4 / 22 = 0.1818

Step 4. Divide to obtain elasticity, then take the absolute value for classification.

E = −0.2222 / 0.1818 = −1.222

|E| = 1.22

Step 5. Classify the result.

Since |E| > 1, demand is elastic — quantity demanded responds more than proportionally to the price change.

Answer: Price elasticity ≈ 1.22 (in absolute value); demand is elastic.

SUBJECT 5 — CALCULUS I

Question 5: Finding and Classifying a Critical Point

Find the critical point of f(x) = x² − 6x + 5 and determine whether it is a minimum or a maximum.

Step 1. Differentiate the function using the power rule.

f(x) = x² − 6x + 5

f′(x) = 2x − 6

Step 2. Set the first derivative equal to zero and solve for x to locate critical points.

2x − 6 = 0

2x = 6 → x = 3

Step 3. Use the second derivative test to classify the point.

f″(x) = 2

Since f″(x) = 2 > 0, the curve is concave up everywhere, so x = 3 is a minimum.

Step 4. Find the y-coordinate by substituting x = 3 back into f(x).

f(3) = (3)² − 6(3) + 5 = 9 − 18 + 5 = −4

Answer: The function has a minimum at the point (3, −4).