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Factor the numerator as (x − 2)(x + 2) and cancel the common factor (x − 2) for x ≠ 2, leaving x + 2. Substituting x = 2 into the simplified expression gives 4. The original quotient has a removable discontinuity at x = 2, but the two-sided limit still exists.
The chain rule states that d/dx f(g(x)) = f′(g(x))·g′(x). The outer function is sine, whose derivative is cosine, and the inner function 3x has derivative 3. Multiplying these together gives 3cos(3x).
For a 2×2 matrix [[a, b], [c, d]] the determinant is ad − bc. Substituting the entries gives (2)(3) − (1)(5) = 6 − 5 = 1.
Expand term by term and use i² = −1: (3)(1) + (3)(−4i) + (2i)(1) + (2i)(−4i) = 3 − 12i + 2i + 8. Combining the real parts (3 + 8) and the imaginary parts (−12i + 2i) gives 11 − 10i.
The law of cosines gives c² = a² + b² − 2ab·cos C. Substituting a = 5, b = 7, and C = 60° yields c² = 25 + 49 − 2(5)(7)(0.5) = 39, so c = √39 ≈ 6.24.
The antiderivative of 3x² is x³, since the derivative of x³ is 3x². Evaluating x³ at the upper limit 2 and subtracting its value at the lower limit 0 gives 8 − 0 = 8.
Integration by parts with u = x and dv = eˣ dx gives ∫x·eˣ dx = x·eˣ − eˣ = (x − 1)eˣ. Evaluating this antiderivative at x = 1 gives 0, and at x = 0 gives −1, so the integral equals 0 − (−1) = 1.
Work equals the dot product F·d, which is the sum of the componentwise products. Here 6 − 8 − 5 = −7 joules, meaning the force removes energy along this displacement.
The binomial probability is C(n, k)·pᵏ·(1 − p)^(n − k). With n = 5, k = 2, and p = 0.2, this is 10 × 0.04 × 0.512 = 0.2048.
Standardize the value: z = (60 − 50)/10 = 1. From the standard normal table, the cumulative probability P(Z < 1) is approximately 0.8413.
A p-series ∑1/nᵖ converges exactly when p > 1. Of the choices, only 1/n³ has an exponent greater than 1, so it is the only convergent series listed.
The equation is separable: dy/y = 2x dx, so ln y = x² + C. Applying the initial condition gives C = ln 3, hence y = 3e^(x²).
Differentiating first with respect to y, treating x as a constant, gives ∂f/∂y = 3x²y². Differentiating that result with respect to x gives 6xy².
On the interval from 0 to 1 the line lies above the parabola, so the area is ∫₀¹ (x − x²) dx = x²/2 − x³/3 evaluated from 0 to 1. This gives 1/2 − 1/3 = 1/6.
Each half-life multiplies the remaining amount by 1/2. Fifteen years spans three half-lives, so the remaining fraction is (1/2)³ = 1/8.
The second equation gives x = 1 + y. Substituting into the first yields 2(1 + y) + 3y = 7, so 5y = 5 and y = 1, which makes x = 2.
Eigenvalues satisfy λ² − (trace)λ + det = 0. The trace is 7 and the determinant is (4)(3) − (1)(2) = 10, so λ² − 7λ + 10 = 0 factors as (λ − 5)(λ − 2) = 0.
The transform of t is 1/s², and multiplying a function by e^(at) shifts the transform by replacing s with s − a. Applying a = 2 gives 1/(s − 2)², valid for s > 2.
The Maclaurin series for eᵘ is Σ uⁿ/n!. Substituting u = x² gives Σ x^(2n)/n!, so the x⁴ term arises from n = 2 and has coefficient 1/2! = 1/2.
Newton–Raphson uses x₁ = x₀ − f(x₀)/f′(x₀). Here f(1) = −1 and f′(1) = 2, so x₁ = 1 − (−1)/2 = 1.5, which is closer to √2 than the starting guess.
The complement rule states that P(A') = 1 - P(A) for any event A and its complement. Here the complement of 'fails' is 'does not fail', so the probability is 1 - 0.35 = 0.65. The two events together must account for the entire sample space, whose probability is 1.
The expected value is the probability-weighted sum E[X] = sum of x times P(x). Computing gives 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1. Each outcome must be weighted by its own probability rather than treated equally.
For a binomial random variable the variance is np(1 - p). Substituting gives 25(0.2)(0.8) = 4. The mean is np = 5, which is a separate quantity from the variance.
In a right-skewed distribution the long tail of large values pulls the mean toward the tail, while the median stays near the bulk of the data. Consequently the mean is greater than the median whenever the skew is to the right. The reverse ordering occurs for left-skewed data.
A Type I error is a false rejection: the null hypothesis is actually true, but the sample evidence leads the analyst to reject it. Its probability is controlled by the significance level alpha. The complementary mistake, failing to reject a false null, is a Type II error.
For independent events the probability that both occur is the product of their probabilities. Therefore the system works with probability 0.98 times 0.95 = 0.931. Independence is essential here; the probabilities are not added.
Conditional probability is defined as P(A|B) = P(A and B) / P(B). Substituting the given values gives 0.2 / 0.4 = 0.50. Dividing by P(B) restricts attention to the portion of the sample space where B has occurred.
The general addition rule is P(A or B) = P(A) + P(B) - P(A and B), because the intersection is counted twice when the probabilities are added. Substituting gives 0.4 + 0.5 - 0.25 = 0.65. The subtraction is required whether or not the events are independent.
Standardizing gives z = (130 - 100)/15 = 2.00. The standard normal table gives P(Z < 2.00) = 0.9772, so the upper-tail probability is 1 - 0.9772 = 0.0228. The tail area above the z value is the requested probability.
The standard error of the sample mean is sigma divided by the square root of n. With sigma = 12 and n = 36, the square root of n is 6, so the standard error is 12/6 = 2.0. This quantity describes the spread of the sampling distribution of the mean.
For a known sigma the interval is the sample mean plus or minus z times sigma divided by the square root of n. The standard error is 10/5 = 2, and the 95% critical value is 1.96, giving a margin of 3.92. The interval is therefore 60 plus or minus 3.92, or (56.08, 63.92).
The sample mean is 6, and the squared deviations are 16, 4, 0, 4, and 16, which sum to 40. Dividing by n - 1 = 4 gives a sample variance of 10, so the sample standard deviation is the square root of 10, approximately 3.162. Division by n - 1 rather than n makes the estimator unbiased.
The coefficient of determination is the square of the correlation, so r-squared equals 0.80 squared, or 0.64. This means about 64% of the variability in the response variable is accounted for by the fitted linear relationship. The remaining variation is due to factors not captured by the model.
Because order does not matter, the count is the number of combinations C(8,3) = 8! / (3! 5!) = 56. The denominator divided by 3! removes the repeated orderings of each group. Permutations would be used only if the order mattered.
The Poisson probability of exactly k events is e to the minus lambda times lambda to the k divided by k factorial. With lambda = 3 and k = 2, the result is e^-3 times 9 divided by 2, which equals 0.224. The factorial in the denominator accounts for the indistinguishability of the two events.
For an exponential distribution with mean mu, the survival probability beyond time t is e to the minus t divided by mu. With t = 5 and mu = 5 the exponent is -1, giving e^-1 = 0.368. The distribution is memoryless, so the median lifetime is less than the mean.
Bayes' theorem reverses the conditional: P(A | defective) equals the joint probability of A and defective divided by the total defect probability. The numerator is 0.6 times 0.02 = 0.012, and the total is 0.012 + 0.4 times 0.05 = 0.032. The ratio 0.012/0.032 is 0.375.
The margin of error equals z times sigma divided by the square root of n, so n equals the square of z times sigma divided by the margin. Substituting gives (1.96 times 10 divided by 2) squared, which equals 96.04. Sample size must be rounded up, so 97 observations are required.
A p-value is the probability of obtaining a test statistic at least as extreme as the observed one, computed under the assumption that the null hypothesis holds. A value of 0.03 indicates such a result would be unusual under the null, and since 0.03 is below 0.05 the null is rejected. The p-value is not the probability that either hypothesis is true.
In a linear model the slope is the expected change in the response variable for a one-unit increase in the predictor. Here a one-meter increase in x corresponds to an average predicted increase of 3.2 kilograms in y. The constant 12.5 is the predicted value of y when x equals zero.
Simple interest is computed only on the original principal, so total interest equals principal times rate times time: $5,000 x 0.08 x 3 = $1,200. No interest is earned on accrued interest.
Straight-line depreciation spreads the depreciable basis (cost minus salvage value) evenly over the life: ($50,000 - $5,000) / 5 = $9,000 per year. Salvage value must be removed from the basis before dividing.
Simple payback is the initial investment divided by the uniform annual net cash flow: $24,000 / $6,000 = 4.0 years. It ignores the time value of money and any cash flows after payback.
The monthly rate is 12%/12 = 1%, and the effective annual rate is (1 + 0.01)^12 - 1 = 0.1268, or 12.68%. Compounding more often than annually always makes the effective rate exceed the nominal rate.
The single-payment compound amount factor applies: F = P(1 + i)^n = $1,000(1.10)^5 = $1,610.51. The growth includes interest earned on previously accrued interest.
The MARR is the hurdle rate a firm sets as the least return it will accept on an investment, reflecting its cost of capital and risk. A project is economically justified only when its rate of return meets or exceeds the MARR.
Discount the uniform annual returns with the series present worth factor: PW = -$40,000 + $12,000(P/A, 10%, 5) = -$40,000 + $12,000(3.7908) = -$40,000 + $45,490 = $5,490. A positive present worth means the project earns more than the MARR.
For a perpetuity, the capitalized cost equals the annual payment divided by the interest rate: $10,000 / 0.08 = $125,000. The principal is never consumed because earnings cover the payout each year.
The cash flow is a uniform series of $2,000 plus an arithmetic gradient of $500. PW = $2,000(P/A, 10%, 5) + $500(P/G, 10%, 5) = $2,000(3.7908) + $500(6.8618) = $7,582 + $3,431 = $11,013.
Setting PW to zero gives $10,000 = $4,000(P/A, i, 3), so the factor must equal 2.5. Since (P/A, 9%, 3) = 2.531 and (P/A, 10%, 3) = 2.487, interpolating places the rate at about 9.7%.