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Marco S.
Paralegal at a Criminal Defense Firm
Tutor Satisfaction Guarantee
SAT
TutorMe
Question:

At the aquarium, the mean age of all the male seals is $$ 13 $$ years, and the mean age of all of the female seals is $$ 19 $$ years. Which of the following must be true about the mean age $$ m $$ of the combined group of the aquarium's male and female seals? (A) $$ m = 16 $$ (B) $$ m > 16 $$ (C) $$ m < 16 $$ (D) $$ 13 < m < 19 $$

Marco S.
Answer:

Answer: (D) This question, typical of one on the SAT's math section, requires more than simply knowing how to calculate the mean — it tests our understanding of what the mean actually means (pun intended). At first blush, we might calculate the mean of $$ 13 $$ and $$ 19 $$ to find $$ 16 $$ — and, of course, this is the mean of those two numbers. However, if we read our problem again, we understand that there is an amount of male seals at the aquarium and an amount of female seals at the aquarium, but we're not sure exactly how many of each. We know that the mean age of the male seals is $$ 13 $$. If we have seven thousand male seals at the aquarium — and what an aquarium that would be — and they are all aged $$ 13 $$, then the mean of their ages is... $$ 13 $$. Similarly, if we have only two lonely female seals at our aquarium, say, one aged $$ 18 $$ and one $$ 20 $$, then the females' mean age would be $$ 19 $$. It should be clear then, that if we take our thousands of aged $$ 13 $$ male seals and only two older female seals and calculate our mean, it's still going to be very, very, close to $$ 13 $$. This is just a random, extreme example to show that the mean of the combined male and female seals' ages could be anywhere within a range — we just don't know exactly, it would depend on the exact numbers of each gender of seal. Since we can't be sure exactly what the answer is, (A) must be ruled out. Now let's consider the inequalities: does $$ m>16 $$ need to be true, as answer choice (B) would suggest? Well, as our earlier example shows, no, it does not need to be true! We could have very many young male seals and only one or two older female seals, which would bring the average down. Let's just practically work this out to double check, assuming we have $$ 10 $$ male seals who are all aged $$ 13 $$ and $$ 2 $$ female seals who are both aged $$ 19 $$: $$ \frac{(13+13+13+13+13+13+13+13+13+13) + (19+19)}{12} = 14$$ There we go, we have proof that $$ m $$ could be less than $$ 16 $$ : (B) is incorrect. For much of the same reasoning, only if we switched the numbers of male and female seals, we'll see that it's very much possible to have a great many of female seals and relatively few male seals, which would drag the combined mean above $$ 16 $$, so (C) is an incorrect answer choice, too. In the end, all we can really know is that the mean must fall somewhere in between $$ 13 $$ and $$ 19 $$, or (D): $$ 13 < m < 19 $$. Note: how do we know it has to be between $$ 13 $$ and $$ 19 $$? Why can't it be higher? Let's quickly consider this: if we have our male seals, whose mean age is $$ 13 $$, how could adding another group of seals whose mean age is HIGHER than that (the female seals) bring DOWN the mean age of the whole group? It couldn't, it could only bring it higher, and vice versa for the upper limit.

French
TutorMe
Question:

Fill in the blanks of the following sentences, conjugating the given verb in either the passé composé or the imparfait, depending on the context: 1) Hier, il _________ trois pommes. (manger) 2) Est-ce que tu ___________ ce film? (regarder) 3) Avant, elle _____________ les poires; maintenant, je ne les aime plus. (aimer) 4) Jean _________ quand Camille __________. (dormir, arriver)

Marco S.
Answer:

1) This sentence is referring to a point in time specifically: yesterday, he ate three apples. It is a finite event with a clear time frame (yesterday), and as such it should be conjugated with the passé composé: Hier, il a mangé trois pommes. 2) Let's think about the length of time the sentence refers to. The sentence is asking if you have already seen this film — an event that would have a very clear beginning and end. Passé composé again: Est-ce que tu as regardé ce film? 3) Whenever we talk about habits, preferences, or tendencies in the past in French, we use the imperfect! Here, as we're talking about her opinion of pears, we should use the imperfect! Think about it like this: is there an instant, a single event in the past in which she liked pears? Probably not — it was an indefinite period of pear-affinity. Avant, elle aimait les poires; maintenant, elle ne les aime plus. 4) This sentence may seem tricky at first, but let's just do what we've been doing and break it down into the timeframes this sentence references. Jean was sleeping when Camille arrived. To start, it seems clear that Camille's arrival had a clear point in time: it's the single moment she arrived! That calls for passé composé. Looking at Jean and his sleepy self, we can discern that he's still sleeping when Camille arrived, and he was before — it's an event that has continued for an indefinite period of time. That situation requires the imparfait! Jean dormais quand Camille est arrivée. *Note that "arriver" is a verb who's passé composé requires the auxiliary verb "être" — and as such, "arrivée" has an extra "e" to agree with Camille's gender.

Algebra
TutorMe
Question:

The admission price at a local baseball stadium is $3.00 for children and $7.50 for adults. For a certain game, 575 people entered the stadium and $2850 was collected. How many children and how many adults attended?

Marco S.
Answer:

At first glance, you may think that there could be several answers to this question — and you would be right if we only knew ONE total: the total amount of attendees, or the total amount of money collected. However, since we know both, this question can be solved with only one correct answer! In order to do this, we must solve a system of equations — that is, we will express both the amount of people who attended the game and the total amount of money collected as algebraic equations. Let's start by looking at the the total attendance! We know that 575 people attended the game, and that they were either children or adults — meaning that the sum of the total children (which we'll call $$ c $$) and the total adults (for our purposes, $$ a $$) who attended is equal to 575: $$ c+a=575 $$ We also know two other helpful pieces of information: the prices of tickets for children and adults, as well as the total money collected. Each children's ticket brought in $3.00, and each adult's ticket brought in $7.50, for a total of $2850: $$ 3c+7.5a=2850 $$ Now that we have two equations, we can work towards solving for each variable, $$ c $$ and $$ a $$. It's important to note that in both equations we've written, the variables stand for the same number; this means that the $$ c $$ from our first equation is the same value as the $$ c $$ in our second equation. So, what is $$c$$'s value? Let's take a look at the first equation. $$ c+a=575 $$ Solving for $$ c $$, we must subtract $$ a $$ from both sides. $$ c=575-a $$ In other words, the total amount of children who attended the game is equal to 575 minus the amount of adults who attended the game. We have solved for $$ c $$, but we still don't have its actual value yet — just an equivalent expression. In order to find the value of a variable in an algebraic equation, we need to have only one variable in the equation. Luckily, we do have another equation... and we have the variable $$ c $$ expressed only in terms of $$ a $$. Let's take a look and put this together. We know that: $$ 3c+7.5a=2850 $$ And we've just discovered that: $$ c=575-a $$ So we can replace the variable $$ c $$ in that first equation with $$ 575-a $$: $$ 3(575-a)+7.5a=2850 $$ Let's distribute the $$ 3 $$ through that set of parentheses to clean up our equation a bit. Remember, you have to multiply both terms in the parentheses by three, and pay attention to the signs! $$ 1725-3a+7.5a=2850 $$ Now we can see the benefit of solving for $$ c $$ initially. In our most recent equation we've manipulated, we only have one variable — $$ a $$! We can now combine like terms to begin to find its value. Combine the $$ -3a $$ and the $$ 7.5a $$: $$ 1725+4.5a=2850 $$ Now subtract $$ 1725 $$ from both sides of the equation to isolate the $$ a $$: $$ 4.5a=1125 $$ All that's left is to divide both sides by $$ 4.5 $$: $$ a=250 $$ And there we go! We now know, by solving our system of equations, that 250 adults attended the game! All we need to do now is find out how many children attended, which is straightforward from here! We know that $$ c=575-a $$, and we now know the value of $$ a $$, so it's a simple plug-in: $$ c=575-250 $$ $$ c=325 $$ And there you have it: 325 children and 250 adults attended the game!

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