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3 years ago

How does the rotating membership of the FOMC work

Business

2 answers:

Semenov [28]3 years ago

6 0

Rotating membership of the FOMC work:

The Federal Open Market Committee, or FOMC, is the Fed's financial policy-making body. It is answerable for definition of a strategy intended to advance stable costs and monetary development. Basically, the FOMC deals with the country's cash supply.

The leaders of the other Reserve Banks fill the staying four democratic situations on the FOMC on a pivoting premise. The entirety of the Reserve Bank presidents, including the individuals who are not casting a ballot individuals, go to FOMC gatherings, partake in the dialogs, and add to the evaluation of the economy and arrangement choices.

The FOMC plans eight gatherings for every year, one about like clockwork or something like that. The Committee may likewise hold unscheduled gatherings as important to survey monetary and budgetary improvements.

The FOMC issues an arrangement articulation following every ordinary gathering that outlines the Committee's financial viewpoint and the strategy choice at that gathering. The Chairman holds a press preparation after each FOMC meeting to talk about the FOMC's arrangement choices and to give setting to those choices .

xz_007 [3.2K]3 years ago

4 0

Answer:

There are four positions in the FOMC, all of the Reserve Bank presidents, including those who are not voting members, attend FOMC meetings, participate in the discussions, and contribute to the assessment of the economy and policy options.

(I hope this helps! :)

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A bag contains 9 red marbles, 5 white marbles, and 6 blue marbles. You draw 4 marbles out at random, without replacement. Find t

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a) 0.0260

b) 0.0681

Explanation:

The bag contains:

r = 9 (number of red marbles)

w = 5 (number of white marbles)

b = 6 (number of blue marbles)

So, the total number of marbles in the bag at the beginning is:

$n=9+5+6=20$

At the 1st attempt, the probability of choosing a red marble is:

$p(r)=\frac{r}{n}=\frac{9}{20}$ (1)

At the 2nd attempt, the 1st red marble is not placed back, so now the number of marbles is (n-1), while the number of red marbles left is (r-1). So the probability of choosing another red marble at the 2nd attempt is

$p(r)=\frac{r-1}{n-1}=\frac{8}{19}$ (2)

With a similar argument, the probabilities of selecting a red marble in the 3rd and 4th attempt are

$p(r)=\frac{7}{18}$ (3)

$p(r)=\frac{6}{17}$ (4)

Therefore, the probability of drawing 4 red marbles in the first 4 attempts without replacing is:

$p(rrrr)=\frac{9}{20}\frac{8}{19}\frac{7}{18}\frac{6}{17}=\frac{3024}{116280}=0.0260$

At the 1st draw, the probability that the marble is not red is:

$p(r^c)=1-p(r)=1-\frac{9}{20}=\frac{11}{20}$

At the 2nd draw, there are 9 red marbles left and 19 total marble left. So, the probability of NOT drawing a red marble is:

$p(r^c)=1-\frac{9}{19}=\frac{10}{19}$

At the 3rd draw, there are 9 red marbles left and 18 total marbles left. So, the probability of NOT drawing a red marble is:

$p(r^c)=1-\frac{9}{18}=\frac{9}{18}$

Finally, with a similar argument the probability of NOT drawing a red marble at the 4th and last attempt is:

$p(r^c)=\frac{8}{17}$

So, the total probability of drawing 4 non-red marbles in the first 4 attempts is:

$p(r^cr^cr^cr^c)=\frac{11}{20}\frac{10}{19}\frac{9}{18}\frac{8}{17}=\frac{7920}{116,280}=0.0681$

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Answer:

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