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almond37 [142]
3 years ago
13

3.

Mathematics
1 answer:
Vitek1552 [10]3 years ago
8 0
To solve this we are going to use the future value of annuity due formula: FV=(1+ \frac{r}{n} )*P[ \frac{(1+ \frac{r}{n})^{kt}-1 }{ \frac{r}{n} } ]
where
FV is the future value
P is the periodic deposit 
r is the interest rate in decimal form 
n is the number of times the interest is compounded per year
k is the number of deposits per year

We know for our problem that P=420 and t=15. To convert the interest rate to decimal form, we are going to divide the rate by 100%: r= \frac{10}{100} =0.1. Since Ruben makes the deposits every 6 months, k=2. The interest is compounded semiannually, so 2 times per year; therefore, k=2.
Lets replace the values in our formula:

FV=(1+ \frac{r}{n} )*P[ \frac{(1+ \frac{r}{n})^{kt}-1 }{ \frac{r}{n} } ]
FV=(1+ \frac{0.1}{2} )*420[ \frac{(1+ \frac{0.1}{2})^{(2)(15)}-1 }{ \frac{01}{2} } ]
FV=29299.53

We can conclude that the correct answer is <span>$29,299.53</span>
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Step-by-step explanation:

You're finding the perimetre of this rectangle, so have to add each side together to find the total distance he traveled (the perimetre of the rectangle, since he travelled in a rectangle).

3*2 is 6,

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10 + 6 is 16.

Have a wonderful day, my friend.

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3 years ago
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In e
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The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.

Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.

It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let \%R_{A} be the quantity of nonzero elements in R_{A}.

Let C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right).

Parity demands that \%R_{A} and\%R_{B} must equal 2 or 4.

Case 1: \%R_{A}=4 and \%R_B=4. There are \left(\begin{array}{l}3\\ 2\end{array}\right)=3 ways to put 2-1's in R_A, so there are 3 ways.

Case 2: \%R_{A}=2 and \%R_B=4. There are 3 ways to position the -1 in R_A, 2 ways to put the remaining -1 in R_B (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of \%R_{A}=4,\%R_{B}=2 for a complete of 24 ways.

Case 3: \%R_A=\%R_B=2. There are 3 ways to put the -1 in R_{A}. Now, there are two cases on what happens next.

  • The 1 in R_B goes directly under the -1 inR_A. There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in R_C andR_D. (Either the 1 comes first inR_C or the 1 comes first in R_D.)
  • The 1 in R_B doesn't go directly under the -1 in R_A. There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

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Step-by-step explanation:

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Answer:

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