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3 years ago

What is office personnel?

Mathematics

1 answer:

Nana76 [90]3 years ago

5 0

Answer:

Office personnel refers to the office chief, sectional chiefs and assistants that carry out all the administrative as well as clerical functions jointly to archive the objectives of an organization.

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What are the common factors of 60 and 72?

ddd [48]

Answer: 1, 2, 3, 4, 6, 12

Calculation/Explanation:

Common factors are factors that both the numbers, in this case, 60 and 72, have in common. The factors listed above are all factors shared by 60 and 72.

3 0

3 years ago

Please help me with these 2 questions.

Bumek [7]

Answer:

For number one the 3.75 oz container for $9.75 would be the better deal

For number two the answer is 6 1/4

3 0

3 years ago

Find the exact area of the surface obtained by rotating the curve about the x-axis. y = 1 + ex , 0 ≤ x ≤ 9

tekilochka [14]

The surface area is given by

$\displaystyle2\pi\int_0^9(1+e^x)\sqrt{1+e^{2x}}\,\mathrm dx$

since $y=1+e^x\implies y'=e^x$ . To compute the integral, first let

$u=e^x\implies x=\ln u$

so that $\mathrm dx=\frac{\mathrm du}u$ , and the integral becomes

$\displaystyle2\pi\int_1^{e^9}\frac{(1+u)\sqrt{1+u^2}}u\,\mathrm du$

$=\displaystyle2\pi\int_1^{e^9}\left(\frac{\sqrt{1+u^2}}u+\sqrt{1+u^2}\right)\,\mathrm du$

Next, let

$u=\tan t\implies t=\tan^{-1}u$

so that $\mathrm du=\sec^2t\,\mathrm dt$ . Then

$1+u^2=1+\tan^2t=\sec^2t\implies\sqrt{1+u^2}=\sec t$

so the integral becomes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec t}{\tan t}+\sec t\right)\sec^2t\,\mathrm dt$

$=\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec^3t}{\tan t}+\sec^3 t\right)\,\mathrm dt$

Rewrite the integrand with

$\dfrac{\sec^3t}{\tan t}=\dfrac{\sec t\tan t\sec^2t}{\sec^2t-1}$

so that integrating the first term boils down to

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\frac{\sec t\tan t\sec^2t}{\sec^2t-1}\,\mathrm dt=2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\frac{s^2}{s^2-1}\,\mathrm ds$

where we substitute $s=\sec t\implies\mathrm ds=\sec t\tan t\,\mathrm dt$ . Since

$\dfrac{s^2}{s^2-1}=1+\dfrac12\left(\dfrac1{s-1}-\dfrac1{s+1}\right)$

the first term in this integral contributes

$\displaystyle2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\left(1+\frac12\left(\frac1{s-1}-\frac1{s+1}\right)\right)\,\mathrm ds=2\pi\left(s+\frac12\ln\left|\frac{s-1}{s+1}\right|\right)\bigg|_{\sqrt2}^{\sqrt{1+e^{18}}}$

$=2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}$

The second term of the integral contributes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\sec^3t\,\mathrm dt$

The antiderivative of $\sec^3t$ is well-known (enough that I won't derive it here myself):

$\displaystyle\int\sec^3t\,\mathrm dt=\frac12\sec t\tan t+\frac12\ln|\sec t+\tan t|+C$

so this latter integral's contribution is

$\pi\left(\sec t\tan t+\ln|\sec t+\tan t|\right)\bigg|_{\pi/4}^{\tan^{-1}(e^9)}=\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

Then the surface area is

$2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}+\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

$=\boxed{\left((2+e^9)\sqrt{1+e^{18}}-\sqrt2+\ln\dfrac{(e^9+\sqrt{1+e^{18}})(\sqrt{1+e^{18}}-1)}{(1+\sqrt2)(1+\sqrt{1+e^{18}})}\right)\pi}$

4 0

4 years ago

A soda bottling plant fills cans labeled to contain 12 ounces of soda. The filling machine varies and does not fill each can wit

AnnZ [28]

Answer:

$12.1-1.64\frac{0.35}{\sqrt{50}}=12.019$

$12.1+1.64\frac{0.35}{\sqrt{50}}=12.181$

So on this case the 90% confidence interval would be given by (12.019;12.181)

And we can conclude at 90% of confidence that the true mean is between (12.019, 12.181) and since the lower limit is higher than 12 we can conclude that the specification is not satisfied

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".

$\bar X=12.1$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=0.35$ represent the population standard deviation

n=50 represent the sample size

Solution to the problem

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

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

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

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

$12.1-1.64\frac{0.35}{\sqrt{50}}=12.019$

$12.1+1.64\frac{0.35}{\sqrt{50}}=12.181$

So on this case the 90% confidence interval would be given by (12.019;12.181)

And we can conclude at 90% of confidence that the true mean is between (12.019, 12.181) and since the lower limit is higher than 12 we can conclude that the specification is not satisfied

8 0

3 years ago

PLEASEE HELP.!! ILL GIVE BRAINLIEST.!! *EXTRA POINTS* DONT SKIP:((

Cloud [144]

Answer:

okaaaay. this is 10 units to the y axis...

Step-by-step explanation:

8 0

3 years ago

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What is office personnel?​

What is office personnel?