if we take 64 to be the 100%, how much is 6¼% off of it?
![\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 64&100\\ x&6\frac{1}{4} \end{array}\implies \cfrac{64}{x}=\cfrac{100}{6\frac{1}{4}}\implies \cfrac{64}{x}=\cfrac{\frac{100}{1}}{\frac{25}{4}}\implies \cfrac{64}{x}=\cfrac{100}{1}\cdot \cfrac{4}{25} \\\\\\ \cfrac{64}{x}=16\implies 64=16x\implies \cfrac{64}{16}=x\implies 4=x \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{it had}}{64}-\stackrel{\textit{leakage}}{4}\implies \stackrel{\textit{remaining}}{60}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccll%7D%20amount%26%5C%25%5C%5C%20%5Ccline%7B1-2%7D%2064%26100%5C%5C%20x%266%5Cfrac%7B1%7D%7B4%7D%20%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B100%7D%7B6%5Cfrac%7B1%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B%5Cfrac%7B100%7D%7B1%7D%7D%7B%5Cfrac%7B25%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B100%7D%7B1%7D%5Ccdot%20%5Ccfrac%7B4%7D%7B25%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B64%7D%7Bx%7D%3D16%5Cimplies%2064%3D16x%5Cimplies%20%5Ccfrac%7B64%7D%7B16%7D%3Dx%5Cimplies%204%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bit%20had%7D%7D%7B64%7D-%5Cstackrel%7B%5Ctextit%7Bleakage%7D%7D%7B4%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bremaining%7D%7D%7B60%7D)
Answer:
A = 216 cm^2
P = 80 cm
Step-by-step explanation:
l = 12
w = 18
As the object is rectangular, the area is height * width and the perimeter is
2(height + width) as there are two of each side.
12 * 18 = 216 U^2 U = unit of measurement, I would presume for a poster you use cm so its 216 cm^2
12 + 18 = 40
40 * 2 = 80 cm
Answer:
The 95% confidence interval estimate for the population mean force is (1691, 1755).
Step-by-step explanation:
According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally.
The sample selected here is <em>n</em> = 30.
Thus, the sampling distribution of the sample mean will be normal.
Compute the sample mean and standard deviation as follows:

Construct a 95% confidence interval estimate for the population mean force as follows:


Thus, the 95% confidence interval estimate for the population mean force is (1691, 1755).