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Reif L.
Economist & Language Teacher
Tutor Satisfaction Guarantee
Mandarin
TutorMe
Question:

有一天小明從早上就開始在房間打電動,一直到晚餐都沒有出來。媽媽很生氣,就對著小明說:小明你__還在打電動!你__不出來,就不__你吃晚餐了! 請問__裡面應該填那些字才能表達媽媽的憤怒? A. 雖然......還......把 B. 竟然......再......讓 C. 究竟......給......要 D. 難道......在......給

Reif L.
Answer:

The answer is B. For answer A, "雖然" means "even if...", which cannot express the mother's anger when she learns that 小明 is still playing video games. "還" is fine here but is not as a strong statement as "再" in this sentence. "把" is totally wrong and doesn't make sense here. For answer C, "究竟" may be used as a way to express anger in other sentences but not in this one. It is often used with a question such as, "你究竟在搞甚麼?", meaning "(...you have messed up something...) What were you doing exactly?" "給" doesn't make sense, while "讓" is actually correct here. For answer D, "難道" is correct here. "在" is wrong because "在" is used to express a "status or at some place" rather than a "motion" or "repetition", which we use "再", but here the sentence wants to stress that 小明 is "still" not coming out of the bedroom for dinner, so we should use "再" instead. "給" means "to give" or "let someone do something", and here the mother said if 小明 doesn't stop playing video games then she would not "let" him have dinner tonight.

Economics
TutorMe
Question:

Given a cost function: $$c(q)= q^3-4q^2+10q$$, please find its average cost function, ac(q), and marginal cost function, m(q). Please also find a $$q^*\neq\ 0$$ that verifies this relationship: $$at\ some\ q^*, m(q=q^*) = ac(q=q^*)=minimum\ of\ average\ cost$$.

Reif L.
Answer:

1. Find the average cost function and marginal cost function. A. average cost function: $$ac(q) = \frac{c(q)}{q} \ \Rightarrow \ ac(q) = (q^3-4q^2+10q)/q = q^2-4q+10$$ B. marginal cost function: $$m(q) = \frac{\mathrm{d}c(q)}{\mathrm{d}q}\ \Rightarrow \ m(q)= \frac{\mathrm{d}(q^3-4q^2+10q)}{\mathrm{d}q}=3q^2-8q+10$$ 2. Find the $$q^*$$, where $$m(q=q^*) = ac(q=q^*)=minimum\ of\ average\ cost$$. A. $$m(q=q^*)=ac(q=q^*)$$: $$Let\ m(q)=ac(q)\ \Rightarrow\ 3q^2-8q+10 = q^2-4q+10\ \Rightarrow\\ 2q^2-4q=0\ \Rightarrow\ 2q(q-2)=0\ \Rightarrow\ q=0\ or\ 2\\ Since\ q^*\neq0, we\ have\ q^*=2.\\ That \ is,\ m(q=2)=ac(q=2)=6.$$ B. $$ac(q^*)=minimum\ of\ average\ cost:$$ $$Observe\ this\ average\ cost\ function\ and\ we\ can\ find\ that\ it\ is\ a\ 3\ degree\ polynomial\\ with\ a\ minimum\ point\ in\ the\ 1st\ quadrant.\\ Minimize\ the\ average\ cost\ function\ just\ like\ finding\ the\ polynomial's\ minimum\\ point: ac(q)=q^3-4q^2+10q=q(q^2-4q+10)=q[(q^2-4q)+10]\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =q[(q-2)^2-4+10]=q[(q-2)^2+6]\\ The\ minimum\ of \ average\ cost\ occurs\ at\ q=2, which\\ minimum\ of\ average\ cost=ac(q=2)=(2)^2-4(2)+10=6.$$

Statistics
TutorMe
Question:

Researchers claim that more than 60% of 30-year-old people in K city still get birthday gifts from their parents. A recent random sampling of 100 30-year-old people found that 65% of them get birthday gifts from their parents. At $$\alpha=0.05$$, does the random sampling provides enough supporting evidence for the claim?

Reif L.
Answer:

First, this question asks you to do a hypothesis testing. Write out the hypotheses first. $$H_0: \rho\leq0.6$$ and $$H_a: \rho>0.6$$, that is, we are testing whether the claim that "more than 60% of 30-year-old people sill get birthday gifts from their parents." The null hypothesis $$H_0$$ means no, and the alternative hypothesis means at the population level yes. Second, since this is a one sample testing without any further assumption and a known hypothetical value, 60% or 0.6, we should use z test. Let's calculate the z value first. $$z = (\hat{\rho}-\rho)/(\sqrt(\rho(1-\rho)/n)$$. Here, n is the sampling size, which is 100, and $$\rho$$ is 0.6, and $$\hat{\rho}$$ is the sampling result 65% or 0.65. So, we can get the z value: $$z=(0.65-0.6)/\sqrt(0.6(1-0.6)/100)=0.05/\sqrt(0.0.0024)=1.021$$ Lastly, compare this z value to the critical value at $$\alpha=0.05$$. Since we are determining whether the percentage if greater than 60%, we use upper one-tailed z test here. By checking the critical value table for z test, we know that $$z_{0.05}=1.645$$. Now, since $$z=1.021<1.645=z_{0.05}$$, we know it does not fall into the critical zone, where $$z_{0.05}>1.645$$, so that there is not enough supporting evidence to reject the null hypothesis, that is, the claim may be not be true.

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