Solve the equation n/6=12 Is I’m correct?????

The cardboard is 10 by 20 before removing a square from each end. Assuming that the square is x inches wide. Therefore, the 20 in side gets reduced by x inches on both sides, or say it becomes 20 - 2x inches. On the other hand, the 10 in side is also reduced by 2x. The x value we get happens to be the height of the box when the sides are folded up.

Thus the volume V = lbh =

V = (20-2x)*(10-2x)*(x)

V = 4x³ - 60x² + 200x

On differentiating, we have dv/dx to be

dv/dx = 12x² - 120x + 200

Using general formula to find the roots of this equation, we can solve that x = 7.886 and x = 2.113

This roots we got are possible values of x, the square we cut. Since 7.886 * 2 = 15.772 inches, this is more than the 10 inch side, henceforth x = 2.113 inches.

You cut 2.113 inches from each corner to obtain the maximum volume.

The sizes of the cubes are

20 - (2 * 2.113) = 15.774

10 - (2 * 2.113) = 5.774

2.113

The volume of the cube is 15.774 * 5.774 * 2.113 = 192.5 cubic inches.

8 0

3 years ago

A data set includes 103103 body temperatures of healthy adult humans having a mean of 98.998.9degrees°f and a standard deviation

dangina [55]

Answer:

$98.9-2.62\frac{0.65}{\sqrt{103}}=98.73$

$98.9+2.62\frac{0.65}{\sqrt{103}}=99.07$

So on this case the 99% confidence interval would be given by (98.73;99.07)

Step-by-step explanation:

Previous concepts

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".

$\bar X=98.9$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=0.65 represent the sample standard deviation

n=103 represent the sample size

Solution to the problem

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

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=103-1=102$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,102)".And we see that $t_{\alpha/2}=2.62$

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

$98.9-2.62\frac{0.65}{\sqrt{103}}=98.73$

$98.9+2.62\frac{0.65}{\sqrt{103}}=99.07$

So on this case the 99% confidence interval would be given by (98.73;99.07)

6 0

3 years ago

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