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Bogdan [553]
3 years ago
10

Justin invested $7,300 in an account paying an interest rate of 1.5% compounded quarterly. Assuming no deposits or withdrawals a

re made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $8,560?
Mathematics
1 answer:
Liula [17]3 years ago
8 0

Answer:

10.7 years.

Step-by-step explanation:

7,300 * (1 + 0.015)^x = 8,560 <=>

7,300 * 1.015^x = 8,560 <=>

1.015^x = 8,560 / 7,300 <=>

1.015^x = 1.1726 <=>

x = 10.694

To the nearest tenth:

x = 10.7 years

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The Polygons are similar. Find the value of x. ​
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7. x = 10 or -8

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Step-by-step explanation:

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Assume that the helium porosity of coal samples taken from any particular seam is Normally distributed with true standard deviat
riadik2000 [5.3K]

Answer:

a) 4.85-2.33\frac{0.75}{\sqrt{20}}=4.46    

4.85+2.33\frac{0.75}{\sqrt{20}}=5.24    

b) 4.56-2.33\frac{0.75}{\sqrt{16}}=4.12    

4.56+2.33\frac{0.75}{\sqrt{16}}=4.99  

c) n=(\frac{1.960(0.75)}{0.2})^2 =54.02 \approx 55

Step-by-step explanation:

Part a

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The Confidence is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that z_{\alpha/2}=2.33

Now we have everything in order to replace into formula (1):

4.85-2.33\frac{0.75}{\sqrt{20}}=4.46    

4.85+2.33\frac{0.75}{\sqrt{20}}=5.24    

Part b

4.56-2.33\frac{0.75}{\sqrt{16}}=4.12    

4.56+2.33\frac{0.75}{\sqrt{16}}=4.99  

Part c  

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =0.4/2 =0.2  we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, replacing into formula (b) we got:

n=(\frac{1.960(0.75)}{0.2})^2 =54.02 \approx 55

5 0
3 years ago
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