The percentage of people who like either chocolate or vanilla is 25% + 15% - 5% = 35%.

Using inclusion-exclusion, the percentage of people who like either activity is: 25% + 15% - 5% = 35%.

Using inclusion-exclusion, the percentage of people who like either activity is 25% + 35% - 10% = 50%.

Solving 2x + 3 < 11 gives 2x < 8, thus x < 4. The maximum integer value is 3.

From the second equation, x = 10 - 2y. Substituting into the first gives 2(10 - 2y) + 3y = 12, solving gives y = 2, x = 6.

Dividing both sides by 2 gives x = 10/2 = 5.

Dividing both sides by 2 gives x ≡ 2 (mod 3), hence the smallest non-negative solution is 2.

2y - 7 = 13 => 2y = 20 => y = 10.

16 = 2^4 => x + 1 = 4 => x = 3

Since 32 can be expressed as 2^5, we have 2^(x+3) = 2^5, thus x + 3 = 5, leading to x = 2.

If 3 machines produce 150 units, then 6 machines will produce 150 * 2 = 300 units.

Molar volume = total volume / number of moles = 60 L / 3 moles = 20 L/mole.

Volume per mole = total volume / number of moles = 60 L / 3 moles = 20 L/mole.

Molality (m) = moles of solute / kg of solvent; assuming 1 L of water = 1 kg, m = 3 moles / 1 kg = 3 m

Rate of 3 pipes = 1/4 tank/hour. Rate of 1 pipe = 1/12 tank/hour. Rate of 5 pipes = 5/12 tank/hour. Time = 12/5 = 2.4 hours.

Let the number be x. Then, 3/5 * x = 30, so x = 30 * 5/3 = 50.

In a 1:2 ratio, the total parts = 1 + 2 = 3. Alcohol = (1/3) * 30 = 10 liters.

Members from ruling party = 30% of 245 = 0.3 * 245 = 73.5, rounded to 74.

Using m1*T1 + m2*T2 = (m1 + m2)*Tf, we find Tf = (300*25 + 200*75) / (300 + 200) = 50°C.

3x + 2 = 11 => 3x = 9 => x = 3.

Solving gives 3x < 9, thus x < 3.

Add 5 to both sides: 3x = 21. Then divide by 3: x = 7.

x = 12 / 3 = 4.

Dividing both sides by 3 gives x ≡ 3 (mod 12), which means x can be 3.

3x = 9 mod 6 simplifies to x = 3 mod 2, so x = 2.

Since 81 = 3^4, we have 3^(2x) = 3^4, thus 2x = 4, giving x = 2.

The number of arrangements of 4 distinct books is 4! = 24.

The number of arrangements of 4 distinct books is 4! = 24.

The balanced equation is 4Al + 3O2 → 2Al2O3, which shows that 4 moles of Al produce 2 moles of Al2O3.

Molality (m) = moles of solute / kg of solvent = 4 moles / 2 kg = 2 m.