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question:A random sample of 35 US cities was taken to record their growing seasons. The sample mean was found to be 190.7 days, and the population standard deviation was known to be 54.2 days. Construct a 95% confidence interval for the true mean population of the growing season.

answer:We are given the following information: {eq}n= 35,bar X = 190.7,sigma = 54.2{/eq} The critical value is obtained from the standard normal table at a given level of significance (0.05). The critical value is equal to 1.96. A 95% confidence interval for the true mean population of the growing season is calculated as follows. {eq}begin{align*} Pleft( {bar X - {Z_{alpha /2}} times dfrac{sigma }{{sqrt n }} < mu < bar X + {Z_{alpha /2}} times dfrac{sigma }{{sqrt n }}} right) &= 0.95 Pleft( {190.7 - 1.96 times dfrac{{54.2}}{{sqrt {35} }} < mu < 190.7 + 1.96 times dfrac{{54.2}}{{sqrt {35} }}} right) &= 0.95 Pleft( {172.744 < mu < 208.656} right) &= 0.95 end{align*}{/eq} Therefore, we can be 95% confident that the true mean population of the growing season is between 172.744 and 208.656 days.

question:Calculate the difference in the levels of the mercury in a vertical U-tube manometer connected to the pressure tappings at the inlet and throat of a venturi meter. The venturi carries a liquid with a relative density of 0.8, has an inlet diameter of 150 mm, a throat diameter of 75 mm, and a pressure difference of 150 mm above the inlet. The volumetric flow rate is 40 liters/s, and the specific gravity of mercury is 13.56.

answer:Given Data: - Inlet diameter, ( D_A = 150,mm ) - Throat diameter, ( D_B = 75,mm ) - Pressure difference, ( z_B - z_A = 0.15,m ) - Volumetric flow rate, ( Q_a = 0.4,m^3/s ) - Relative density of the liquid, ( text{SG}_{text{liquid}} = 0.8 ) - Specific gravity of mercury, ( text{SG}_{text{mercury}} = 13.56 ) - Coefficient of discharge, ( C_d = 0.96 ) Using the continuity equation for constant discharge: [ frac{V_A}{A_A} = frac{V_B}{A_B} ] [ left(frac{D_B}{D_A}right)^2 = frac{V_B}{4V_A} ] Applying Bernoulli's principle between points A and B: [ frac{p_A}{rho g} + frac{V_A^2}{2g} + z_A = frac{p_B}{rho g} + frac{V_B^2}{2g} + z_B ] [ frac{V_B^2 - frac{V_B^2}{16}}{2g} = frac{p_A - p_B}{rho g} - 0.15 ] Solving for ( V_B ): [ V_B = sqrt{frac{32g}{15}left(frac{p_A - p_B}{rho g} - 0.15right)} ] The equation of actual discharge is: [ Q_a = C_d times V_B times A_B ] [ 0.4 = 0.96 times sqrt{frac{32g}{15}left(frac{p_A - p_B}{rho g} - 0.15right)} times frac{pi D_B^2}{4} ] Solving for ( p_A - p_B ): [ frac{p_A - p_B}{rho g} = 33.82,kN/m^2 ] Let ( h ) be the difference in mercury levels. [ frac{p_A - p_B}{rho g} = h left(frac{text{SG}_{text{mercury}}}{text{SG}_{text{liquid}}} - 1right) ] [ 33.82 times 10^3 = 1000 times 0.8 times 9.81 times 16h ] [ h = 0.26,m ] Therefore, the difference in the levels of the mercury in the two sides of the manometer is ( h = 0.26,m ).

question:1. A 105-g ice cube at 0°C is mixed with 1.0 kg of water initially at 71°C. Determine the final temperature of the water after the ice has completely melted. 2. A 67-g ice cube at 0°C undergoes a transformation, resulting in 60.6 g of water at 100°C, and 6.4 g of steam at 100°C. Calculate the energy input required for this transformation.

answer:1. To find the final temperature, we'll use the principle of calorimetry: Let T be the final temperature, and consider the following: - Mass of ice: ( m_{ice} = 0.105 , kg ) - Mass of water: ( m_w = 1 , kg ) - Specific heat capacity of water: ( c_w = 4200 , J/(kg cdot °C) ) - Latent heat of fusion of ice: ( L_{f} = 334 times 10^{3} , J/kg ) - Initial temperature of water: ( T_2 = 71 , °C ) - Initial temperature of ice: ( T_1 = 0 , °C ) Heat lost by water is equal to the heat gained by ice: [ m_w cdot c_w cdot (T_2 - T) = m_{ice} cdot L_{f} + m_{ice} cdot c_w cdot (T - T_1) ] [ 1 , kg cdot 4200 , J/(kg cdot °C) cdot (71 - T) = 0.105 , kg cdot (334 times 10^{3} , J/kg) + 0.105 , kg cdot 4200 , J/(kg cdot °C) cdot (T - 0) ] [ 298200 - 4200T = 35070 + 441T ] [ 441T + 4200T = 298200 - 35070 ] [ 4641T = 263130 ] [ T = frac{263130}{4641} approx 56.7 , °C ] So, the final temperature is approximately 56.7°C. 2. The energy input (H) is the sum of the energy required to melt the ice, heat the resulting water to 100°C, and vaporize the water to steam: - Mass of ice: ( m_{ice} = 67 , g ) - Mass of steam: ( m_{steam} = 6.4 , g ) - Latent heat of fusion of ice: ( L_{f} = 334 , J/g ) - Latent heat of vaporization of water: ( L_{v} = 2230 , J/g ) [ H = m_{ice} cdot L_{f} + m_{ice} cdot c_w cdot (100 - 0) + m_{steam} cdot L_{v} ] [ H = 67 , g cdot 334 , J/g + 67 , g cdot 4.2 , J/(g cdot °C) cdot 100 + 6.4 , g cdot 2230 , J/g ] [ H = 22518 + 2802 + 14208 ] [ H = 40528 , J ] Therefore, 40,528 Joules of energy was added to accomplish the transformation.

question:Compute the inverse of the following matrix: left( begin{array}{ccc} 4 & 0 & -1 2 & -3 & 2 5 & 1 & 0 end{array} right)

answer:The inverse of the given matrix is: left( begin{array}{ccc} frac{2}{25} & frac{1}{25} & frac{3}{25} -frac{2}{5} & -frac{1}{5} & frac{2}{5} -frac{17}{25} & frac{4}{25} & frac{12}{25} end{array} right) This has been calculated using the standard method for finding the inverse of a matrix, which involves calculating the determinant, cofactor matrices, and then transposing the cofactor matrix divided by the determinant. The determinant of the original matrix is -25, which is non-zero, ensuring that the inverse exists.