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victus00 [196]
3 years ago
5

A bar of metal is cooling from 1000°C to room temperature, 24°C. The temperature, H, of the bar t minutes after it starts coolin

g is given, in oc, by H = 24 + 976e-0.1t
(a) Find the temperature of the bar at the end of one hour. (Round your answer to the nearest degree.)
(b) Find the average value of the temperature over the first hour. (Round your answer to the nearest degree.)
(c) Is your answer to part (b) greater or smaller than the average of the temperatures at the beginning and the end of the hour? The answer to part (b) isecthan the average of the temperatures at the beginning and the end of the hour, because the graph is concave
Mathematics
1 answer:
motikmotik3 years ago
7 0

Answer:

ggfddv3=4

Step-by-step explanation:

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lana66690 [7]

Answer:

a) n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83  

And rounded up we have that n=4906

b) For this case since we don't have prior info we need to use as estimatro for the proportion \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724  

And rounded up we have that n=6724

Step-by-step explanation:

We need to remember that the confidence interval for the true proportion is given by :  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

Part a

The estimated proportion for this case is \hat p =0.76

Our interval is at 90% of confidence, and the significance level is given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. The critical values for this case are:

z_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

The margin of error desired is given ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

Replacing we got:

n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83  

And rounded up we have that n=4906

Part b

For this case since we don't have prior info we need to use as estimatro for the proportion \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724  

And rounded up we have that n=6724

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Step-by-step explanation:

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