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3 years ago

14

if the federal reserve deceases the reserve rate from 5% to 2%, how does this affect the amount of money that would result becau

se of fractional reserve banking from an initial deposit into a bank of 25,000

1 answer:

muminat3 years ago

5 0

25,000÷0.05
=500,000

25,000÷0.02
=1,250,000
increased by
1,250,000−500,000
=750,000

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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

Sedaia [141]

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 in $R_A$ . There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in $R_C$ and $R_D$ . (Either the 1 comes first in $R_C$ 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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2 years ago

Read 2 more answers

Please please help, need this done by 1:45!!!!

GarryVolchara [31]

Answer:

option B. MB/AM=NC/AN

Step-by-step explanation:

we know that

The <u><em>Triangle Proportionality Theorem</em></u> states that if a line is parallel to one side of a triangle and it intersects the other two sides, then it divides those sides proportionally

In this problem

MN is parallel to BC

MN intersect AC and divide into AN and NC

MN intersect AB and divide into AM and MB

so

Applying the Triangle Proportionality Theorem

$\frac{AN}{NC}=\frac{AM}{MB}$

Rewrite

$\frac{MB}{AM}=\frac{NC}{AN}$

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3 years ago

Which set of line segments could create a right triangle?

ser-zykov [4K]

Answer:

15,36,39

Step-by-step explanation:

A right triangle will satisfy Pythagoras Theorem:

c² = a² + b²; where c is the longest side.

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3 years ago

0.0000056 rewritten as a single digit<br> times a power of ten becomes:

Masteriza [31]

Answer:

5.6× 10^ -6

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3 years ago

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the n

Ivanshal [37]

Answer:

<u><em>a) The probability that exactly 4 flights are on time is equal to 0.0313</em></u>

<u><em></em></u>

<u><em>b) The probability that at most 3 flights are on time is equal to 0.0293</em></u>

<u><em></em></u>

<u><em>c) The probability that at least 8 flights are on time is equal to 0.00586</em></u>

Step-by-step explanation:

The question posted is incomplete. This is the complete question:

<em>United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures. </em>

<em>a) The probability that exactly 4 flights are on time is = </em>

<em>b) The probability that at most 3 flights are on time is = </em>

<em>c)The probability that at least 8 flights are on time is =</em>

<h2>Solution to the problem</h2>

<u><em>a) Probability that exactly 4 flights are on time</em></u>

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

$P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}$

Where $_nC_x$ is the number of different combinations of x success in n trials.

$_nC_x=\frac{x!}{n!(n-x)!}$

Hence,

$P(X=4)=_9C_4\cdot (0.5)^4\cdot (0.5)^{5}$

$_9C_4=\frac{4!}{9!(9-4)!}=126$

$P(X=4)=126\cdot (0.5)^4\cdot (0.5)^{5}=0.03125$

<em><u>b) Probability that at most 3 flights are on time</u></em>

The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=0)=(0.5)^9=0.00195313$ . . . (the probability that all are not on time)

$P(X=1)=_9C_1(0.5)^1(0.5)^8=9(0.5)^1(0.5)^8=0.00390625$

$P(X=2)=_9C_2(0.5)^2(0.5)^7=36(0.5)^2(0.5)^7=0.0078125$

$P(X=3)= _9C_3(0.5)^3(0.5)^6=84(0.5)^3(0.5)^6=0.015625$

$P(X\leq 3)=0.00195313+0.00390625+0.0078125+0.015625=0.02929688\\\\ P(X\leq 3) \approx 0.0293$

<em><u>c) Probability that at least 8 flights are on time </u></em>

That at least 8 flights are on time is the same that at most 1 is not on time.

That is, 1 or 0 flights are not on time.

Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.

$P(Y=0)=_0C_9(0.5)^0(0.5)^9=0.00195313\\ \\ P(Y=1)=_1C_9(0.5)^1(0.5)^8=0.0039065\\ \\ P(Y=0)+P(Y=1)=0.00585938\approx 0.00586$

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4 years ago