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miskamm [114]
3 years ago
6

The Federal Reserve conducts a $15 million open-market purchase of government bonds. If the required reserve ratio is 20 percent

, the largest possible increase in the money supply that could result is $ ................. million, and the smallest possible increase is $ .....................million.
Business
1 answer:
Soloha48 [4]3 years ago
8 0

Answer:

$200 million

$30 million

Explanation:

When the requiredreserce ratio is 15 percent or 0.15 , then the money multiplier is (1 / required reserve ratio) or (1/0.15 = 0.67)

Now, change in money supply = money multiplier * open market purchase of government bonds.

Here , the Federal Reserve a $30 million open market purchase Of govemment bonds.

As a result of this;

Money Supply increases by (6.7 * $30 million) = $200 million.

This is the maximum amount the money supply could Increase.

Now, if the bank holds. $30 million as excess reserves, then money supply could increase by as much as $30 million. This is the smallest amount themoney supply could increase.

So, If the required reserve ratio is 15 percent the largest possible increase in the money supply that could result is $200 million- and the smallest possible increase is $30 million.

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Which of the following statements about prices and inflation is not​ correct? A. Inflation represents a general rise in prices o
nevsk [136]

Answer: Prices generally increase at the same rate across most periods of time.

Explanation:

Inflation means a rapid rise in the price of commodities in a market, and it is normally as a result of scarcity of products or excess flow of money in an economy. Prices on the other hand do not always increase generally, as price could reduce or remain the same overtime.

7 0
3 years ago
Read 2 more answers
What number should be included in the following data so that the median is 5.2? 4,7,11,5,3​
ruslelena [56]

Answer:

5.4

Explanation:

To find the median, we need to arrange the numbers in numerical order.

Hence, here we have 3, 4, 5, 7, 11.

Given that the question says we should look for numbers that would be included in the following data so that the median is 5.2. Hence, we need to add the number to the set.

Therefore the total number of the set will be 6 making it an even number. Hence we have to add the two numbers in the middle and divide them by 2.

Since the median is 5.2, hence we multiply it by 2 = 10.4

Therefore, we less 5 from 10.4, we have 5.4

To check we now have a set of numbers 3, 4, 5, 5.4, 7, 11

Where 5 + 5.4 = 10.4, then divided by 2 = 5.2 as median.

4 0
3 years ago
Calculate the payback period for the following investment: Invest ($10,000). Cash flows in Yr1: 1,000; Yr2: 3,000; Yr3: 3,000; Y
mihalych1998 [28]

Answer:

4 years

Explanation:

The computation of the payback period is shown below:

In the payback, we analyze in how many years the invested amount is recovered

In year 0 = -$10,000

In year 1 = $1,000

In year 2 = $3,000

In year 3 = $3,000

In year 4 = $3,000

In year 5 = $100,000

In year 6 = $250,000

If we sum the first 4 year cash inflows than it would be $10,000

And, the initial investment is also $10,000

So, in 4 years, the investment amount is recovered

8 0
3 years ago
A job can be done with Machine A that costs $12,500 and has annual end-of-year maintenance costs of $5000; its salvage value aft
sdas [7]

Answer and Explanation :

As per the data given in the question,

Present value = Amount ÷ (1 + r)^n

Machine A

Year           Amount        Discount Factor      PV

1                  $5,000           1.05                  $4,761.90

2                $5,000                                     $4,535.15

3               $5,000                                      $4,319.19

Total                                                          $13,616.24

Now

Present value of salvage value =$2,000 ÷ 1.05^3 = $1,727.68

Present worth of Machine A is

= -$12,500 - $13,616.24 + $1,727.68

= -$24,388.56

Similarly Present worth of Machine B = -$15,000 - $4,000 ÷ 1.05 -$4,000 ÷ (1.05)^2 - $4,000 ÷ 1.05^3 - $4,000 ÷ 1.05^4 + $1,500 ÷ 1.05^4

=-$24,658.94

Based on the comparison between Machine A and Machine B

Machine A is better because it has higher present worth

Annual worth:

For machine A = -$12,500(A/PA,5%,3) -$5,000+$2,000(A/F,5%,3)

=-$12,500 × 0.367 - $5,000 + $2,000 × 0.317

= -$8,953.5

For Machine B:

=  -$15,000(A/P,5%,4) - $4,000 + $1,500(A/F,5%,4)

= -$7,882.16

Based on the comparison between Machine A and Machine B

Machine B is better because it has higher annual worth

Capitalized cost:

Machine A :

= -$12,500+$2,000(P/F,5%,3) - $5,000 ÷ 0.05

=  -$12,500 + $2,000 × 0.86 - $5,000 ÷ 0.05

= -$110,772

Machine B :

=-$15,000(P/F,5%,4) - $4000 ÷ 0.05

=-$15,000 × 0.82 - $4,000 ÷ 0.05

= -$93,765.9

Based on the comparison between Machine A and Machine B

Machine B is better because it has lower capitalized cost

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3 years ago
In what way, if any, is pharming different from baiting?
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C I think is the answer
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