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The $x/\ln (x)$ approximation yields an estimated count of 3.1 trillion primes below 100 trillion. Compared to the actual count of 3.2 trillion, this is over 99.99% accurate.

Conditions: $x = 100$,000,000,000,000.; The formula used is #primes(x) ~ $x/\ln (x)$.

The $x/\ln (x)$ approximation yields an estimated count of 3.1 trillion primes below 100 trillion. Compared to the actual count of 3.2 trillion, this is over 99.99% accurate.

Conditions: $x = 100$,000,000,000,000.; The formula used is #primes(x) ~ $x/\ln (x)$.

The present value is $55,839.48. This is calculated by discounting the future value of $100,000 back 10 years using the formula$PV = $\frac{FV}{(1+r)^n}$$, where$$r = 0.06$$and$$n = 10$$.

Conditions: The future value FV is $100,000.; The annual interest rate r is 6% (0.06).; The number of periods n is 10 years.; Compounding is annual.

The present value is $55,839.48. This is calculated by discounting the future value of $100,000 back 10 years using the formula$PV = $\frac{FV}{(1+r)^n}$$, where$$r = 0.06$$and$$n = 10$$.

Conditions: The future value FV is $100,000.; The annual interest rate r is 6% (0.06).; The number of periods n is 10 years.; Compounding is annual.

The video demonstrates that a smaller population size (e.g., 50) leads to worse solutions compared to a larger population (e.g., 100). Specifically, the run with population size 50 produced routes with self-crossings and a longer final path length (4823) than the previous run with population size $100 (4296)$.

Conditions: Observed with fixed iterations = 100 and the same displayed interface settings.; The conclusion is based on the shown runs rather than a formal proof.; Comparison is made against the previously shown larger-population run.

The video demonstrates that a smaller population size (e.g., 50) leads to worse solutions compared to a larger population (e.g., 100). Specifically, the run with population size 50 produced routes with self-crossings and a longer final path length (4823) than the previous run with population size $100 (4296)$.

Conditions: Observed with fixed iterations = 100 and the same displayed interface settings.; The conclusion is based on the shown runs rather than a formal proof.; Comparison is made against the previously shown larger-population run.

The expected value of -$0.50 should be interpreted as a long-run average loss per ticket, not as a guaranteed result for a single play. In any individual game, the player will either win $499 or lose $1; the -$0.50 figure represents the mean outcome if the game were played many times.

Conditions: The expected value has been calculated as -$0.50.; The context involves understanding the meaning of expectation in probability.

The expected value of -$0.50 should be interpreted as a long-run average loss per ticket, not as a guaranteed result for a single play. In any individual game, the player will either win $499 or lose $1; the -$0.50 figure represents the mean outcome if the game were played many times.

Conditions: The expected value has been calculated as -$0.50.; The context involves understanding the meaning of expectation in probability.

The definite integral of the piecewise-linear function from 0 to 7 equals 10.5. This result is obtained by summing the signed areas of the three sub-intervals created by splitting at the breakpoints: 4.5, 6, and 0.

Conditions: The function is piecewise-linear.; The integration interval is from 0 to 7.; The graph consists of regions with signed areas 4.5, 6, and 0.

The definite integral of the piecewise-linear function from 0 to 7 equals 10.5. This result is obtained by summing the signed areas of the three sub-intervals created by splitting at the breakpoints: 4.5, 6, and 0.

Conditions: The function is piecewise-linear.; The integration interval is from 0 to 7.; The graph consists of regions with signed areas 4.5, 6, and 0.

5 has no inverse modulo 10 because the greatest common divisor of 5 and 10 is 5, which is not 1. According to the existence theorem, an integer has an inverse modulo m if and only if their gcd is 1.

Conditions: Modulus is 10.; Number whose inverse is requested: 5.

5 has no inverse modulo 10 because the greatest common divisor of 5 and 10 is 5, which is not 1. According to the existence theorem, an integer has an inverse modulo m if and only if their gcd is 1.

Conditions: Modulus is 10.; Number whose inverse is requested: 5.

Multiplying a fraction by $\frac{5}{5}$ or $\frac{4}{4}$ keeps it equivalent because each of these factors equals 1. Any nonzero number divided by itself equals 1, and multiplying any quantity by 1 does not change its value.

Conditions: The factor used is a fraction whose numerator and denominator are equal and nonzero.; The original fraction is being rewritten to obtain a specific common denominator.

Multiplying a fraction by $\frac{5}{5}$ or $\frac{4}{4}$ keeps it equivalent because each of these factors equals 1. Any nonzero number divided by itself equals 1, and multiplying any quantity by 1 does not change its value.

Conditions: The factor used is a fraction whose numerator and denominator are equal and nonzero.; The original fraction is being rewritten to obtain a specific common denominator.

To compute $\gcd(10,15)$, repeatedly apply the identity $\gcd(a,b)=\gcd(b,a\bmod b)$ until the second argument becomes zero. Starting from $\gcd(10,15)$, the sequence proceeds as $\gcd(15,10)$, then $\gcd(10,5)$, and finally $\gcd(5,0)$. Since the second argument is zero, the algorithm terminates and returns the first argument, 5, as the greatest common divisor.

Conditions: The initial pair is (10, 15).; The algorithm uses the Euclidean step $\gcd(a,b)=\gcd(b,a\bmod b)$.; The process stops when the second argument is zero.

To compute $\gcd(10,15)$, repeatedly apply the identity $\gcd(a,b)=\gcd(b,a\bmod b)$ until the second argument becomes zero. Starting from $\gcd(10,15)$, the sequence proceeds as $\gcd(15,10)$, then $\gcd(10,5)$, and finally $\gcd(5,0)$. Since the second argument is zero, the algorithm terminates and returns the first argument, 5, as the greatest common divisor.

Conditions: The initial pair is (10, 15).; The algorithm uses the Euclidean step $\gcd(a,b)=\gcd(b,a\bmod b)$.; The process stops when the second argument is zero.