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Anit [1.1K]
3 years ago
8

Trella alcala deposited 1,950 in a new credit union savings account on the first of the quarter the principal earns 4.25 percent

interest compound quarterly she made no other deposits or withdrawals
What was the amount in her account at the end of 6 months
What is the compound interest
Mathematics
2 answers:
4vir4ik [10]3 years ago
8 0

Answer:

The amount in her account at the end of 6 months is $1991.58.

The compound interest is $41.58.

Step-by-step explanation:

The compound interest formula is given by:

A = P(1 + \frac{r}{n})^{nt}

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

In this exercise, we have:

A = 1950, n = 3, r = 0.0425

What was the amount in her account at the end of 6 months:

This is a when t = 0.5 years.

So

A = P(1 + \frac{r}{n})^{nt}

A = 1950(1 + \frac{0.0425}{3})^{3*0.5} = 1991.58

The amount in her account at the end of 6 months is $1991.58.

What is the compound interest?

The compound interest is the amount subtracted by the principal. So 1991.58-1950 = $41.58.

mrs_skeptik [129]3 years ago
5 0
Amount in compound interest = p(1 + r/t)^nt where p is the initial deposit, r = rate, t = number of compunding in a period and n = period.

Here, Amount after 6 months (0.5 year) = 1,950(1 + (4.25/100)/4)^(0.5 x 4) = 1,950(1 + 0.0425/4)^2 = 1,950(1 + 0.010625)^2 = 1,950(1.010625)^2 = 1,950(1.0213629) = $1,991.66

Compound interest = Amount - principal (initial deposit) = $1,991.66 - $1,950 = $41.66
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Answer:

x = 0.529 and y = 3.882

Step-by-step explanation:

The given equations are :

y=\dfrac{5}{3}x+3\ ......(1)\\\\y=\dfrc{1}{3}x-3\ .......(2)

We need to solve the equations by graphing.

The attached figure shows the solution of the given equations. We can see that, the intersecting points are :

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We average n=45 samples to get \bar{x}.


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\sigma = 0.68 / \sqrt{45} = 0.101


Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains \mu.


Our interval takes the form of ( \bar{x} - z \sigma, \bar{x} + z \sigma ) as \bar{x} is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".


Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.


Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.


With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.


We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is


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in other words a margin of error of


\pm 1.65(.101) = \pm 0.167 dollars


That's around plus or minus 17 cents.




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