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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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Points A,B, and C are collinear. Points M and N are the midpoints of segments AB and AC. Prove that BC = 2MN
irinina [24]
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2)|AM| = |MB| = x
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Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles. C = 71°, a = 27, c = 26
guajiro [1.7K]

We have

SinC/ c  = Sin A / a

Sin 71/ 26 = Sin A / 27

Sin A  = 27 Sin 71 /  26  = about .982

So°

Sin-1(.982)  = A  = 79. 08°

Then angle B = 180 - 71 - 79.08 = 29.92°

And b is given by

b/sin29.92  = 26/sin 71

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But A could also be an obtuse angle = 180 - 79.08 =  100.92°

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B = 180 - 71 - 100.92 = 8.08°

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According to a recent​ survey, the population distribution of number of years of education for​ self-employed individuals in a c
creativ13 [48]

Answer:

a) X: number of years of education

b) Sample mean = 13.5, Sample standard deviation = 0.4

c) Sample mean = 13.5, Sample standard deviation = 0.2

d) Decrease the sample standard deviation

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 13.5 years

Standard deviation,σ = 2.8 years

a) random variable X

X: number of years of education

Central limit theorem:

If large random samples are drawn from population with mean \mu and standard deviation \sigma, then the distribution of sample mean will be normally distributed with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

b) mean and the standard for a random sample of size 49

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c) mean and the standard for a random sample of size 196

\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{196}} = 0.2

d) Effect of increasing n

As the sample size increases, the standard error that is the sample standard deviation decreases. Thus, quadrupling sample size will half the standard deviation.

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