Can some one help me with this, I can't get it, and i'll give brainliest, 5 star, and all the good stuff.

Let R be the region bounded by the following curves. Let S be the solid generated when R is revolved about the given axis. Ifpo

steposvetlana [31]

The two curves $y=x$ and $y=x^{1/15}$ intersect at $x=0$ and $x=1$ , with $x^{1/15}\ge x$ over $0\le x\le1$ .

Washer method

$\displaystyle\pi\int_0^1\left(\left(x^{1/15}\right)^2-x^2\right)\,\mathrm dx=\pi\int_0^1\left(x^{2/15}-x^2\right)\,\mathrm dx=\pi\left(\frac{15}{17}-\frac13\right)=\boxed{\frac{28\pi}{51}}$

Shell method

We have $y=x^{1/15}\implies x=y^{15}$ . The curves $x=y$ and $x=y^{15}$ intersect at $y=0$ and $y=1$ , with $y\ge y^{15}$ over $0\le y\le1$ .

$\displaystyle2\pi\int_0^1y\left(y-y^{15}\right)\,\mathrm dy=2\pi\int_0^1\left(y^2-y^{16}\right)\,\mathrm dy=2\pi\left(\frac13-\frac1{17}\right)=\boxed{\frac{28\pi}{51}}$

6 0

3 years ago

A manufacturer considers his production process to be out of control when defects exceed 3%. In a random sample of 100 items, th

pochemuha

Answer:

The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.

Step-by-step explanation:

A manufacturer considers his production process to be out of control when defects exceed 3%.

At the null hypothesis, we test if the production process is in control, that is, the defective proportion is of 3% or less. So

$H_0: p \leq 0.03$

At the alternate hypothesis, we test if the production process is out of control, that is, the defective proportion exceeds 3%. So

$H_1: p > 0.03$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.03 is tested at the null hypothesis

This means that $\mu = 0.03, \sigma = \sqrt{0.03*0.97}$

In a random sample of 100 items, the defect rate is 4%.

This means that $n = 100, X = 0.04$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.04 - 0.03}{\frac{\sqrt{0.03*0.97}}{\sqrt{100}}}$

$z = 0.59$

P-value of the test

The p-value of the test is the probability of finding a sample proportion above 0.04, which is 1 subtracted by the p-value of z = 0.59.

Looking at the z-table, z = 0.59 has a p-value of 0.7224

1 - 0.7224 = 0.2776

The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.

4 0

3 years ago

How is 16.666666 . . . written as a fraction

AVprozaik [17]

1/3 ANS

hope this helps and i hope u ace it

5 0

3 years ago

Nd the 52nd term and the term named in the problem <br>40, 140, 240, 340, ...<br>Find a 32

Galina-37 [17]

Answer:

5140 and 3140