answer:To find the product of x and y, we can multiply the numerators and denominators of the fractions: x cdot y = left(-frac{6-22 i}{e}right) cdot left(frac{23+23 i}{e}right) = frac{(-6-22 i)(23+23 i)}{e^2} = frac{-138+156 i-506 i+529}{e^2} = frac{391-350 i}{e^2} = -frac{644-368 i}{e^2} Therefore, the product of x and y is -frac{644-368 i}{e^2}. The answer is -frac{644-368 i}{e^2}

answer:Using the given side lengths and angle, we can calculate the following: - Area: A = frac{1}{2} times a times b times sin(C) where a = 7.7, b = 9.7, and sin(C) = sin(44 {}^{circ}). After computation, A = 25.94 square units. - Circumradius (R): R = frac{abc}{4A} where c is the third side length, which can be found using the Law of Cosines: c^2 = a^2 + b^2 - 2abcos(C). After finding c and then R, we get R = 4.88 units. - Inradius (r): r = frac{A}{s} where s is the semiperimeter, s = frac{a + b + c}{2}. Calculating s, we find s = 12.09 units, and thus r = 2.15 units. - Semiperimeter (already given): s = 12.09 units. The triangle's area is 25.94 square units, the circumradius is 4.88 units, the inradius is 2.15 units, and the semiperimeter is 12.09 units.

answer:First, assume a 100g sample of the compound. This gives: 40.0 g of carbon 6.7 g of hydrogen 53.3 g of oxygen Moles of carbon: 40.0/12.0107 = 3.33 moles Moles of hydrogen: 6.7/1.00794 = 6.65 moles Moles of oxygen: 53.3/15.9994 = 3.33 moles Divide all moles by the smallest value (3.33 moles): 3.33/3.33 = 1 6.65/3.33 = 2 3.33/3.33 = 1 Thus, the empirical formula is CH₂O.

answer:The correct answer is (a). 0.3156. According to the Central Limit Theorem, the sampling distribution of the mean is normally distributed even if the parent population is not, provided the sample size is large (typically greater than 30): {eq}bar{X} sim N(mu, sigma^2/n) {/eq} Here, we need to find: {eq}P(bar{X} < 170) {/eq} Given: - Population mean, {eq}mu = 172 text{ pounds} {/eq} - Population standard deviation, {eq}sigma = 29 text{ pounds} {/eq} - Sample size, {eq}n = 50 {/eq} First, calculate the z-score using the formula for a sample mean: {eq}begin{align*} z &= frac{bar{X} - mu}{frac{sigma}{sqrt{n}}} &= frac{170 - 172}{frac{29}{sqrt{50}}} &= -0.48 end{align*} {/eq} Using a standard normal distribution table or calculator, we find the probability corresponding to a z-score of -0.48: {eq}begin{align*} P(bar{X} < 170) &= P(Z < -0.48) &= 0.3156 end{align*} {/eq} Therefore, the probability that the mean weight of a random sample of 50 men is less than 170 pounds is 31.56%.