GingerAle

Here’s the math portion of the problem:

\(P(B)=0.3 \tiny \text{ Read as “The probability that a child (from the sample set) likes (B)rownies" }\\ P(C)=0.5 \tiny \text{ The probability that a child likes (C)upcakes}\\ P(B\cap C)=0.15 \tiny \text{ (The intersection of these probabilities )}\\ \\ P(B|C)=\dfrac{P(B\cap C)}{P(C)}=\dfrac{0.15}{0.5}=0.3 \tiny \text { Read as “The probability that a child likes brownies, given that s/he likes cupcakes.}\\ P(C|B)=\dfrac{P(B\cap C)}{P(B)}=\dfrac{0.15}{0.3}=0.5 \tiny \text { The probability that a child likes cupcakes, given that s/he likes brownies.}\\ \small \text {Note that }P(B|C)=P(B) \small \text { and }P(C|B)=P(C)\\ \)

With this, you can logically determine which statement is true.

Post your response on here, clearly explaining your logic.

If you do not respond, I will not answer any more of your questions.

Gavin, your post is a menagerie of atrocious logic and mathematics. I’m greatly tempted to troll you.

In these kinds of questions, it is important to calculate the total quantities poured into the cup. The solution is simple and straightforward.  The changing ratios are irrelevant for this question. (You didn’t calculate the ratios correctly either. Ex. In #3, the milk went from 1/3 to 1/9 just by drinking it. How can that happen in a homogenized mixture?).

This question starts with an 8 oz. cup of coffee. No more coffee is added to the cup.

Milk is added in increments of 1/6 cup, 1/3 cup, and 1/2 cup.  This totals to one (1) cup.

From this, it’s easy to tell that the drinker consumes one (1) cup of coffee and one (1) cup of milk. Equal amounts.   

GA

There is a “Chicken McNugget’s Theorem,” but there should not be one.

The real reason is the “cute” name “Chicken McNugget’s Theorem.”  If it’s “cute” then more will remember it and use it. Cuteness in general, and anthropomorphic cuteness in particular, is the byword for educating children and adults too.  We might not understand it unless a blòódy dancing McNugget explains it to us.

Solving two variables in a single equation is usually a trial and error process. However, in this case, you can easily narrow the process to a few guesses using logic.

The (i) and (j) values are integers; there are no fractional or decimal components in them.

Because the sum of integers is very low, it reasonable to assume the (i,,j) exponents are low—less than 10.

To solve, start by assuming a central value (between 1 and 10) for (j), and then use logarithms to isolate for (i).  

Because (2) is one of the factors, its resolution will be its multiples 2,4,8,16,32,64,128,256, ... ect. Seeing one of these indicates you’ve found the solution set.

\((( 2^i-1)*\dfrac{(3^j - 1)}{(2)} = 600 \leftarrow \tiny \text{ The (2) and (1) are common to all equations.}\\ ( 2^i) = \dfrac{1200}{(3^j - 1)} +1\leftarrow \tiny \text{ Leave it in this form to streamline the solution search.}\\ \text {Example: assume (j) = 6 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(3^6 - 1)} +1\\ ( 2^i) = \dfrac{150}{91} +1\\ Log_2( 2^i) = Log_2 \left( \dfrac {150}{91} +1\right) \small \text{ Note the fraction; This is not the solution.}\\ \text {Example: assume (j) = 5 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(5^5 - 1)} +1\\ ( 2^i) = \dfrac{600}{121} +1\\ Log_2( 2^i) = Log_2 \left( \dfrac {600}{121} +1\right) \small \text{ Again, note the fraction; This is not the solution.}\\ \text {Example: assume (j) = 4 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(5^4 - 1)} +1\\ ( 2^i) = 15 +1\\ Log_2( 2^i) = Log_2 \left( 16\right) \small \text{ Note the integer 16, corresponding to i=4; This is the solution.}\\ \)

The solution is j=4 and i=4.  Subtract (1) from each and use these as the exponents for the primes 2, and 3 to find the number (N) for which 600 is the sum of its divisors.

There are similar techniques for (N) that have three or more unique prime factors.

GA

Edit: Corrected to indicate base (2) log.

Solution:

\(E= Z_c \sqrt{\frac{\hat{p}(1-\hat{p})}{N}} \tiny \text{ (E is the margin of error. } Z_c \text{ is the critical value for the confidence level. } \hat{p} \text { is portion of interest.)}\\ \small \text {Arrange and solve for (N) to find the minimum sample size} \\ N= \left( \frac {Z_c /100}{E} \right)^{2} \hat{p}\left(1-\hat{p}\right) \\ \small \text {For 95% confidence }Z_c \small \text { } = 1.96\\ N= \left( \frac {1.96}{0.07} \right)^{2} \ * \ 0.68\left(1-0.68 \right) \; \ = \; \lceil{170.60}\rceil \Leftarrow \small \text {Round up to next integer}\\ \text {Minimum sample size }\mathbf {171}\\\)

Additional information:

For this question, the population is assumed to be above 20,000 (because no population is specified), well above what would be needed for a (US) high school population that averages 860 students per school, with the highest at 1500. For schools with these populations, the sample minimums calculate to 145 and 154 persons respectively. At 171 samples, the margin of error for these schools drops to 6.25% and 6.50% respectively.   

GA

TMC, your post is almost magical. It’s to the point, concise, and well articulated.  Not to forget, your answer is correct.  The most magical part is that a “kid” can make a monkey out of an old turkey (Mr. BB). If it’s not magical, it’s certainly a mystical gift, and though unintentional, it’s very funny. 

GA

Solution:

A divisor of a number is an integer that divides the number with a remainder of zero. All integers greater than one (1) have at least two (2 )divisors. If a number has only two divisors then it is a prime number.

\(\small \text{To find the total number of divisors of a number, factor out the primes }\\ \small \text{and add one (1) to the exponent of each prime. }\\ (4^3) \cdot (5^4) \cdot (6^2) = (2^8) \cdot(3^2)\cdot(5^4)\\ (8+1)(2+1)(4+1) = \mathbf {135} \text { divisors for } (4^3) \cdot (5^4) \cdot (6^2)\\ \text{ }\\ \text{Additional notes: }\\ (2^8) \cdot(3^2)\cdot(5^4)\\ \small \text{All factors will be in this form } (2^x) \cdot(3^y)\cdot(5^z) \\ \small \text {where x,y,z are} \geq 0 \text{ and} \leq \{8,2,4\} \small \text{ the highest prime factor exponents of the number itself.} \\ \small \text {As such, the calculation of all individual divisors may be found by iterating (stepping through) each }\\ \small \text{prime’s exponent from zero to the maximum and multiplying by the other primes. }\\ \sum_\limits {x=0}^{8} \sum_\limits {y=0}^{2}\sum_\limits {z=0}^{4} ((2^x)(3^y)(5^z)) =5,188,183 \text{ (sum of divisors) }\\ \)

GA