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question:Factor the following quadratic: -2 x^2 - 26 x

answer:To factor the quadratic -2 x^2 - 26 x, we can first factor out a common factor of -2 x from both terms: -2 x^2 - 26 x = -2 x (x + 13) Therefore, the factors of -2 x^2 - 26 x are -2 x and (x + 13). The answer is -2 x (x + 13)

question:An object is accelerated from an initial velocity of 18 m/s at a rate of 4 m/s^2. What is the magnitude of the velocity of the object at the end of 5 seconds?

answer:The velocity of an object that starts from an initial velocity of 18 m/s and accelerates at a rate of 4 m/s^2 after 5 seconds can be calculated using the following equation: v = u + at where: v is the final velocity u is the initial velocity a is the acceleration t is the time Substituting the given values into the equation, we get: v = 18 m/s + 4 m/s^2 * 5 s v = 18 m/s + 20 m/s v = 38 m/s Therefore, the magnitude of the velocity of the object at the end of 5 seconds is 38 m/s.

question:A particle's displacement in centimeters is given by the equation y = 8 sin(2πft). If the frequency f is 20 Hz, calculate the displacement and velocity of the particle at times t = 0.01 s and t = 0.07 s.

answer:Given data: Displacement equation: y = 8 sin(2πft) Frequency: f = 20 Hz Displacement at t = 0.01 s: y = 8 sin(2πf * t) y = 8 sin(2π * 20 * 0.01) y = 0.175 m Velocity at t = 0.01 s: v = dy/dt v = d/dt (8 sin(2πft)) v = 8 cos(2πft) * 2πf v = 8 cos(2π * 20 * 0.01) * 2π * 20 v = 1005.06 m/s Displacement at t = 0.07 s: y = 8 sin(2πf * t) y = 8 sin(2π * 20 * 0.07) y = 1.22 m Velocity at t = 0.07 s: v = dy/dt v = d/dt (8 sin(2πft)) v = 8 cos(2πft) * 2πf v = 8 cos(2π * 20 * 0.07) * 2π * 20 v = 993.485 m/s Therefore, at t = 0.01 s, the displacement is 0.175 m and the velocity is 1005.06 m/s. At t = 0.07 s, the displacement is 1.22 m and the velocity is 993.485 m/s.

question:A study monitored the {eq}CO_2 {/eq} levels in 57 kitchens with gas cooking appliances over a one-week period. The sample mean {eq}CO_2 {/eq} level (in ppm) was 654.16, with a sample standard deviation of 164.04. (a) Calculate a 95% (two-sided) confidence interval for the true average {eq}CO_2 {/eq} level in the population of all homes. Interpret the interval. (b) If the investigators had guessed a standard deviation of 165 ppm, how many kitchens would they need to monitor to obtain a 95% confidence interval with a width of 56 ppm?

answer:(a) To find the 95% confidence interval, we use the formula: {eq}CI = overline{x} pm z_{alpha/2} cdot frac{s}{sqrt{n}} {/eq} Given: - {eq}n = 57 {/eq} - {eq}overline{x} = 654.16 {/eq} - {eq}s = 164.04 {/eq} - {eq}1 - alpha = 0.95, text{so} ; alpha = 0.05 ; text{and} ; z_{alpha/2} = 1.96 {/eq} Calculating the confidence interval: {eq}CI = 654.16 pm 1.96 cdot frac{164.04}{sqrt{57}} {/eq} {eq}CI = 654.16 pm 1.96 cdot frac{164.04}{7.54} {/eq} {eq}CI = 654.16 pm 1.96 cdot 21.73 {/eq} {eq}CI = 654.16 pm 42.59 {/eq} {eq}CI = (654.16 - 42.59, 654.16 + 42.59) {/eq} {eq}CI = (611.57, 696.75) {/eq} Interpretation: B. We are 95% confident that this interval, from 611.57 ppm to 696.75 ppm, contains the true population mean. (b) To find the sample size needed, we use the formula: {eq}n = left( frac{z_{alpha/2} cdot sigma}{E} right)^2 {/eq} Given: - {eq}z_{alpha/2} = 1.96 {/eq} - {eq}sigma = 165 ; text{(estimated)} {/eq} - {eq}E = 56 ; text{(desired width)} {/eq} Calculating the sample size: {eq}n = left( frac{1.96 cdot 165}{56} right)^2 {/eq} {eq}n = left( frac{323.4}{56} right)^2 {/eq} {eq}n = (5.775)^2 {/eq} {eq}n = 33.35 {/eq} Since sample sizes must be whole numbers, round up to: Sample Size: 34 Hence, the investigators would need to monitor 34 kitchens to achieve the desired interval width and confidence level.