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Numerical Answers (workings/explanations are after the numerical answers.)
Question 4: ; away from the starting point
Note to students: You can and should be using GC here.
By comparing coefficients of real and imaginary parts,
Using GC, other roots are and .
From the graph, we can observe that each rectangle has a constant width of and the height of each rectangle is given by the coordinate. Also note that the same of rectangles here is an underestimation of the area under the curve from to .
Sum of rectangles
Sum of rectangles
Sum of rectangles
Actual area under the graph
Explanation (if you are interested): The rectangles in (a) touches the curve on the left corners while the rectangles in (b) is meant to touch the curve on the right corners. Thus, we simply need to “shift” one term to the right.
Note: There are many possible answers here, a decreasing function should be good. My choice came up to me as the easiest. I will verify it.
using the GC.
Consider , then f is not defined when since is a vertical asymptote. Thus, .
(b)(ii) For functions to be self-inverse, we need both functions to have the same rule, domain and range. Observe that has a horizontal asymptote at . Then .
Observe that , however the function will not be a 1-1 function if f is a horizontal line, thus .
(b)(iii) Since we have that it is self-inverse, then .
Consider , this implies that Suzie is on her 17th lap. Thus she is swimming away from her starting point.
Let be the score of the spinner.
Using GC, .
(a) Number of ways
(b) Number of ways
(c) Required probability
(a) Manager should carry out a 1-tail test since he wishes to test if the population mean life span is greater than 20 000 miles.
Let be the population mean life span of tyres.
Let be the null hypothesis.
Let be the alternative hypothesis.
Unbiased estimates of population mean miles
Unbiased estimates of population variance
Note that if you did not multiply, i.e., did in thousand miles. It is perfectly fine, but you should note to state the instead, in order to be coherent for the value.
Under , since is sufficiently large, by central limit theorem,
Since , we do not reject at 5% significance level and conclude with insufficient evidence that the mean life span of the tyre is more than 20000 miles.
As the distribution of the tyre is unknown initially, we will need to take a sufficiently large sample size that is greater than 30 in order to approximate the sample mean life span to a normal distribution by central limit theorem. Thus a sample size of 15 will be inappropriate.
The given product moment correlation coefficient suggests a strong positive linear correlation between the bivariate data. The scatter diagram also suggest that as increases, increases proportionately. Thus a linear model will be appropriate.
Both models suggest that there is a strong positive linear correlation between the bivariate data. However, the product moment correlation coefficient for model in (b) is closer to 1 and suggests a stronger linear correlation as compared to that for model in (a). Realistically, as time passes, the mass should increase decreasingly and thus the logarithm model is more appropriate.
(a) Let be the mass of the seat, in kg.
Using GC, and
(b) Let be the mass of the legs, in kg.
(e) Let and be the diameters of the holes in the seats and diameters of the holes of the legs, respectively, in mm.
(a)(i) The light fittings should be selected such that he obtains a random sample, i.e., each light fitting has an equal chance of being selected, independently of each other.
(a)(ii) There are several ways to solve this, the best would be to use the GC and calculate using the frequency functions. The probability here is akin to proportion of light fitting that is faulty out of the 150 days of 100 samples each day. So .
Note that expected number need not be a whole number.
(b)(i) The event that the heating elements for electric ovens is found to be faulty is independent of another heating elements for electric ovens. The probability that heating elements for electric ovens is found to be faulty is constant thoughout the sample.
Let be the number of days with more than 3 elements found to be faulty, out of 5 days.
Let be the number of faulty elements found, out of 400 elements.