If the price of a purse went from $45 to $54, what was the percent of the change?

First, we must count the total number of coins in the cash register. Do not change the value of these coins. Simply take the numbers listed in the problem and add them up. Your process for completing this step should look something like this: 10+11+13+6 equals 40. With this being said, there are 40 coins in the cash register.

Next, we must remember the values of the two coins the question is asking for: pennies and dimes. Keep in mind that, out of all 40 coins, 10 of them are pennies and 13 of them are dimes.

Then, we write the fractions used to find the answer to this problem. In the question, the cashier (hypothetically) draws a penny first, and then a dime. This means a penny is drawn before a dime is drawn. (Although it seems like a miniscule detail, this knowledge is actually something extremely important to remember. This small piece of information can actually change the entire answer if not used properly.)

In this problem, the cashier would choose a single penny out of the 10 in contrast to the total 40 coins. Since there are 10 pennies overall, and 40 coins overall, the probability of choosing a penny first is 10/40.

After we complete the probability of choosing a penny before all else, we must find the probability of choosing a dime after choosing a penny. Since the dime will be chosen after the penny is chosen, one coin will be gone: the chosen penny. So, the total amount of coins within the cash register changes (or more specifically decreases (by 1)) from 40 to 39. However, the amount of dimes, 13, stays the same. With this information in mind, we can write a fraction of 13/39 for the probability of choosing a dime after choosing a penny.

Finally, now that we have both probability fractions, we can figure out the answer to the question. In order to do this, we multiply and simplify both fractions. 10/40*13/39 is 130/1560. 130 ÷ 130 is 1. 1560 ÷ 130 is 12. 1 ÷ 12 can also be written as 1/12. Therefore, the answer to this problem is G) 1/12.

<u>(The paragraph and author's note below are technically optional thought processes and information, but read them if you wish. Although I would recommend reading further, considering another way to solve the question is mentioned, it is entirely you (the reader's) choice.)</u>

You can also check this answer through the other/alternate way of solving this problem: by simplifying both probability fractions before you multiply them. 10/40 can be simplified to 1/4. 13/39 can be simplified to 1/3. After doing this, we multiply both simplified probability fractions in order to find, or in this case check, the answer. 1/4*1/3 is 1/12. 1/12 is the same fraction as the one present in choice G. This is the same answer as the one we previously got. This enforces the idea that G) 1/12 is indeed the answer to this question.

<em> A/N (Author's Note): Although it took quite a while to jot down, this was truly a blast to write and explain! I hope this helped those that needed/wanted it. Have a wonderful week reader(s)!</em>

8 0

2 years ago

Find the roots of x^4-6x^2+8

slavikrds [6]

In my Knowledge

x^4-6x^2+8=0

8 0

3 years ago

A fair coin is tossed repeatedly with results Y0, Y1, Y2, . . . that are 0 or 1 with probability 1/2 each. For n ≥ 1 let Xn = Yn

Gekata [30.6K]

Answer:

False. See te explanation an counter example below.

Step-by-step explanation:

For this case we need to find:

$P(X_{n+1} = | X_n =i, X_{n-1}=i') =P(X_{n+1}=j |X_n =i)$ for all $i,i',j$ and for $X_n$ in the Markov Chain assumed. If we proof this then we have a Markov Chain

For example if we assume that $j=2, i=1, i'=0$ then we have this:

$P(X_{n+1} = | X_n =i, X_{n-1}=i') =\frac{1}{2}$

Because we can only have $j=2, i=1, i'=0$ if we have this:

$Y_{n+1}=1 , Y_n= 1, Y_{n-1}=0, Y_{n-2}=0$ , from definition given $X_n = Y_n + Y_{n-1}$

With $i=1, i'=0$ we have that $Y_n =1 , Y_{n-1}=0, Y_{n-2}=0$

So based on these conditions $Y_{n+1}$ would be 1 with probability 1/2 from the definition.

If we find a counter example when the probability is not satisfied we can proof that we don't have a Markov Chain.

Let's assume that $j=2, i=1, i'=2$ for this case in order to satisfy the definition then $Y_n =0, Y_{n-1}=1, Y_{n-2}=1$

But on this case that means $X_{n+1}\neq 2$ and on this case the probability $P(X_{n+1}=j| X_n =i, X_{n-1}=i')= 0$ , so we have a counter example and we have that:

$P(X_{n+1} =j| X_n =i, X_{n-1}=i') \neq P(X_{n+1} =j | X_n =i)$ for all $i,i', j$ so then we can conclude that we don't have a Markov chain for this case.

6 0

2 years ago