Prove that there is solution to 3tanx = 2 + sinx on the interval [0, pi/4].
Proving the existence of a solution to the above equation on [0, pi/4] is the same as proving that the function f(x) = 3tanx - (2 + sinx) has a root on [0, pi/4]. That is, proving there exists some c in (0,pi/4) that gives 3tanc - (2 + sinc) = 0. Note that f(x) is continuous for all x in our interval. Next, notice that f(0) = -2 < 0 f(pi/4) = 3(1) - (2 + 1/sqrt(2)) > 0 Thus, it follows that f(0) < 0 < f(pi/4). So, by the Intermediate Value Theorem, there indeed exists a c in (0, pi/4) that produces 3tanc - (2 + sinc) = 0 Or, put as the question was phrased 3tanc = 2 + sinc
Assumed we have a standard 52-card deck of playing cards. If 5 cards are dealt to you, what's the probability that you get a hand with 2 kings, 2 aces, and a queen?
This is a classical probability problem, so we use the ratio form P = (desired outcome) / (total number of outcomes) Note that there are four types of each card in a deck. Thus, if our hand is to have 2 kings, 2 aces, and queen, the number of combinations is (4 choose 2)(4 choose 2)(4 choose 1); this is our desired outcome. The total number of outcomes is (52 choose 5). It follows that our probability is: (4 choose 2)^2 * (4 choose 1) / (52 choose 5)
Suppose SuperWheel is a company which claims they only produce 16 defective tires, on average, per year. We collect a sample of size (n) 31 from the company and find the sample average (xbar) of 30 defective tires and a standard deviation (sd) of 13. Assuming the data come from a normal population, can we reject SuperWheel's claim?
Here's the information we can pick out: n = 31 mu0 = 16 (null hypothesis population mean) xbar = 30 sd = 13 Our data is normal (assumed) Since our data is normal and n > 29, we can construct the follow two-sided z-test: H0: mu = 16 H1: mu doesn't equal 16 alpha = 0.05 Reject H0 if |T| > z(alpha/2) where T = sqrt(n)*(xbar - mu0)/sd. We chose alpha = 0.05 since the significance level was not stated in the problem (we could've also another value, but 0.05 or 0.01 are the standards when not specified). Note that z(alpha/2) = z(0.025) = 1.96 (we look at z-score table for this value) and after plugging into a calculator, we get T = 5.49. Conclusion: Since 5.49 > 1.96, it follows that we reject the null hypothesis. That is, there is strong evidence that SuperWheel doesn't create 16 defective wheels per year.