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mr_godi [17]
3 years ago
15

From 1960 to 2010, a certain money stock measure was growing at the rate of approximately 43e(1/2)x billion dollars per decade,

where x is the number of decades since 1950. Find the total increase in the money stock measure from 1960 to 2010. (Round your answer to the nearest billion dollars.)
Mathematics
1 answer:
natulia [17]3 years ago
6 0

Answer:

1222 billion dollars.

Step-by-step explanation:

To find the total increase from 1960 to 2010, we need to find the growth of each decade and sum them all:

In the period 1960-1970, we have x = 1, and the growth is:

y(1) = 43e(1/2) = 70.895

In the period 1970-1980, we have x = 2, and the growth is:

y(2) = 43e(2/2) = 116.8861

The growth in the following 3 periods are:

y(3) = 43e(3/2) =192.7126

y(4) = 43e(4/2) = 317.7294

y(5) = 43e(5/2) =523.8472

So the total growth in the period 1960 - 2010 is:

Total = y(1) + y(2) + y(3) + y(4) + y(5)

Total = 70.895 + 116.8861+192.726+317.7294+523.8472

Total = 1222.08\ billion\ dollars

Rounding to the nearest billion dollars, we have a total of 1222 billion dollars.

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How would I do this problem?
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If it's a geometric sequence then:

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3 years ago
If the distribution is really N ( 5.43 , 0.54 ) , N(5.43,0.54), what proportion of observations would be less than 5.05 ? 5.05?
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Answer:

P(X

And we can use the following excel code to find the answer: "=NORM.DIST(-0.704,0,1,TRUE)"

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable of interest for a population, and for this case we know the distribution for X is given by:

X \sim N(5.43,0.54)  

Where \mu=5.43 and \sigma=0.54

We are interested on this probability

P(X

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X

And we can find this probability on this way:

P(Z

And we can use the following excel code to find the answer: "=NORM.DIST(-0.704,0,1,TRUE)"

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