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nexus9112 [7]
3 years ago
5

Kimmys savings account has a balance of $76.23 in june. By September, her balance is 5 times as much as her June balance. Betwee

n September and December, Kimmy deposits a total of $87.83 into her account. If she does not withdraw any money from her account, what should Kimmys balance be in December
Mathematics
2 answers:
erik [133]3 years ago
5 0

The <em><u>correct answer</u></em> is:

$468.98.

Explanation:

The account balance in June is 76.23. Since the value in September is 5 times that, the September balance is 5(76.23) = 381.15.

She deposits 87.83 between September and December; this makes the total 381.15+87.83 = $468.98.

Andru [333]3 years ago
4 0
If she doesn't withdraw (spend) any money she would have a total of 468.98$
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Suppose that the researchers wanted to estimate the mean reaction time to within 6 msec with 95% confidence. Using the sample st
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Answer:

a) The 95% confidence interval would be given by (509.592;550.308)  

b) n=523 rounded up to the nearest integer  

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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".  

\bar X=530 represent the sample mean for the sample  

\mu population mean

s=70 represent the sample standard deviation  

n=48 represent the sample size (variable of interest)  

Confidence =95% or 0.995

Part a

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

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

The degrees of freedom are df=n-1=48-1=47

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,47)".And we see that z_{\alpha/2}=2.01  

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

530-2.01\frac{70}{\sqrt{48}}=509.692  

530+2.01\frac{70}{\sqrt{48}}=550.308  

So on this case the 95% confidence interval would be given by (509.592;550.308)  

Part b

The margin of error is given by this formula:  

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

Assuming that \hat \sigma =s

And on this case we have that ME =6msec, and 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 is provided, z_{\alpha/2}=1.96, replacing into formula (b) we got:  

n=(\frac{1.96(70)}{6})^2 =522.88 \approx 523  

So the answer for this case would be n=523 rounded up to the nearest integer  

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