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topjm [15]
3 years ago
11

HELP ME NOW PLEASE NOW ADGAFDKD;LHJ!!!!!! PLEASE AND THANK YOU :)))))

Mathematics
1 answer:
Eddi Din [679]3 years ago
7 0

I got D.

There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:

1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator

1 - .1131 - .2005 = .6864

Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.

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We know that,

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Here,

FV\ of\ annuity=2,500,000,\\\\P=?,\\\\r = 9\%\ annually=\frac{9}{12}\%\ monthly=\frac{9}{1200}\ monthly\\\\n=45\ years=45\times 12=540\ months

Putting the values,

\Rightarrow 2500000=P[\frac{(1+\frac{9}{1200})^{540}-1}{{\frac{9}{1200}}}]

\Rightarrow P=\frac{2500000}{[\frac{(1+\frac{9}{1200})^{540}-1}{{\frac{9}{1200}}}]}

\Rightarrow P=\frac{2500000}{\frac{56.5365-1}{0.0075}}

\Rightarrow P=\frac{2500000}{\frac{55.5365}{0.0075}}

\Rightarrow P=337.62


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