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Law Incorporation [45]
3 years ago
5

If 1 euro is priced at $1.25 and if 1 euro will also buy 88 Japanese yen (€1 = ¥88), in equilibrium, with no arbitrage opportuni

ties, how much is the cross rate between the yen and the dollar (yen per dollar exchange rate)?
Mathematics
1 answer:
Alborosie3 years ago
5 0

Answer:

So the exchange rate is that each dollar is worth 70.4 yens.

Step-by-step explanation:

We have that:

1 euro is priced at $1.25

1 euro will also buy 88 Japanese yen

This also means that:

$1.25 is priced at 88 Japanese yen

How much is the cross rate between the yen and the dollar (yen per dollar exchange rate)?

How much yens is 1 dollar?

This can be solved by a simple rule of three.

$1.25 - 88 yen

$1 - x yen

1.25x = 88

x = \frac{88}{1.25}

x = 70.4

So the exchange rate is that each dollar is worth 70.4 yens.

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29/50 or 0.58 is the answer
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Evaluate the expression 7x2y, when x = 4 and y = 2. and just gave me the plain answer
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224

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Read 2 more answers
Choose all that give the correct expression for the quantity described. The difference of nine times a number x and the quotient
ohaa [14]

Step-by-step explanation:

We are to get the expression for the following statements;

1) The difference of nine times a number x and the quotient of that number and 5.

The product of nine and a number x is expressed as;

=9 \times x\\= 9x

The quotient of that number and 5.

= \frac{x}{5}

The difference between both expression;

9x - \frac{x}{5}

Hence, the difference of nine times a number x and the quotient of that number and 5 is expressed as 9x - \frac{x}{5}

2) Eight more than the quotient of twelve and a number n

Quotient of twelve and a number n is expressed as:

\frac{n}{12}

Eight more than the resulting function is;

\frac{n}{12}+8

Hence eight more than the quotient of twelve and a number n is expressed as \frac{n}{12}+8

3) The product of a number and the quantity 'six minus the number' plus the quotient of eight and the number.

Let the number be x:

six minus the number is expressed as;

6-x

product of a number x and the quantity 'six minus the number is;

x(6-x)

quotient of eight and the number is;

\frac{8}{x}

Taking the resulting sum of the last two expression

x(6-x) + 8x

Hence the product of a number and the quantity 'six minus the number' plus the quotient of eight and the number is expressed as;

x(6-x) + 8x

4) Sum of three consecutive even integers. 2x + (2x + 2) + (2x + 4).

Let the first even number be 2x

The consecutive even numbers are gotten by adding 2 to the preceding number. The two consecutive even integers are 2x+2 and 2x+2+2

the sum of three consecutive even integers is expressed as;

= 2x +(2x+2)+(2x+2+2)\\=  2x+(2x+2)+(2x+4)

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3 years ago
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