answer:At 25 degrees Celsius, the concentration of hydronium ions ({eq}[H_3O^+] {/eq}) in a solution with a pH of 3.19 can be calculated using the following equation: rm{left [ H_3O^+ right ]=10^{-pH}=10^{-3.19}=boxed{6.46times 10^{-4}M} } Given the relationship {eq}left [ H_3O^+ right ]left [ OH^- right ]=K_w {/eq}, where {eq}K_w {/eq} is the ion product constant of water ({eq}1.0 times 10^{-14} {/eq} at 25 degrees Celsius), the concentration of hydroxide ions ({eq}[OH^-] {/eq}) can be calculated as follows: rm{left [ OH^- right ]=frac{K_w}{left [ H_3O^+ right ]}=frac{1.0times 10^{-14}}{6.46times 10^{-4}M}=boxed{1.55times10^{-11}M}} Therefore, in a solution with a pH of 3.19 at 25 degrees Celsius, the concentration of hydronium ions is {eq}6.46times 10^{-4}M {/eq} and the concentration of hydroxide ions is {eq}1.55times10^{-11}M {/eq}.

answer:To find the p-value associated with a given z-score, we can use a standard normal distribution table or a calculator. In this case, z = -1.52 corresponds to a p-value of approximately 0.0643. This means that the probability of observing a z-score as extreme as -1.52 or more extreme, assuming the null hypothesis is true, is 0.0643 or 6.43%.

answer:To find the equation of the plane, we can use the normal vector obtained by taking the cross product of two vectors formed by the given points. Let's denote the points as P_1(-a, -b, c), P_2(0, 0, -d), and P_3(e, -f, -g), where a, b, c, d, e, f, and g are the coordinates. First, find two vectors: vec{P_1P_2} = (0, 0, -d) - left(-frac{7}{2}, -2, frac{9}{2}right) = left(frac{7}{2}, 2, -frac{9}{2} - frac{5}{2}right) = left(frac{7}{2}, 2, -7right) vec{P_1P_3} = left(frac{1}{2}, -5, -1right) - left(-frac{7}{2}, -2, frac{9}{2}right) = left(frac{1}{2} + frac{7}{2}, -5 + 2, -1 - frac{9}{2}right) = (4, -3, -frac{11}{2}right) Now, find the normal vector vec{n} by taking their cross product: vec{n} = vec{P_1P_2} times vec{P_1P_3} = left|begin{array}{ccc} mathbf{i} & mathbf{j} & mathbf{k} frac{7}{2} & 2 & -7 4 & -3 & -frac{11}{2} end{array}right| After calculating the cross product, we get the normal vector vec{n} = left(-frac{7}{2}, -14, frac{19}{2}right). The equation of the plane is given by: vec{n} cdot (vec{r} - vec{r_0}) = 0 Where vec{r} is any point on the plane, vec{r_0} is a known point on the plane (e.g., P_2(0, 0, -frac{5}{2})), and cdot denotes the dot product. Substituting the components and simplifying, we get: -frac{7}{2}(x - 0) - 14(y - 0) + frac{19}{2}(z + frac{5}{2}) = 0 -frac{7}{2}x - 14y + frac{19}{2}z + frac{95}{4} = 0 -7x - 28y + 19z = -frac{95}{4} 196x + 784y - 76z + 95 = 0 This equation represents the plane in question. However, it can be simplified by dividing through by 196 to obtain a more standard form: x + 4y - frac{19z}{196} = -frac{95}{784} Rounded to three significant figures, we get: 1.00x + 4.00y - 0.10z = -0.125 So, the equation of the plane is approximately: x + 4y - 0.10z = -0.125 Alternatively, we can express the equation in the form Ax + By + Cz + D = 0: 1.00x + 4.00y - 0.10z + 0.125 = 0 Converting it to standard form, we get: 120x + 480y - 10z + 31 = 0 This is a more compact form, but it is not as precise as the previous form. For this exercise, we'll stick with the more precise form: x + 4y - 0.10z = -0.125

answer:The simplified form of the expression is -frac{2357947691 left(cos left(frac{504}{45}right)+i sin left(frac{504}{45}right)right)}{81 sqrt{3}}. Note: The original answer has a mistake in the argument of the trigonometric functions in the simplified form. The angle should be multiplied by 9, which is frac{56 times 9}{45} = frac{504}{45}.