Help plz due today i dont want to fail!

Convert the integral below to polar coordinates and evaluate the integral.

White raven [17]

I suppose the integral is

$\displaystyle\int_0^5\int_y^{\sqrt{25-y^2}} xy\,\mathrm dx\,\mathrm dy$

The integration region corresponds to a sector of a cirlce with radius 5 subtended by a central angle of π/4 rad. We can capture this region in polar coordinates by the set

$R=\left\{(r,\theta)\mid0\le r\le5,0\le\theta\le\dfrac\pi4\right\}$

Then $x=r\cos\theta$ , $y=r\sin\theta$ , and $\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$ . So the integral becomes

$\displaystyle\iint_Rxy\,\mathrm dx\,\mathrm dy=\int_0^{\pi/4}\int_0^5r^3\sin\theta\cos\theta\,\mathrm dr\,\mathrm d\theta=\frac{625}{16}$

7 0

3 years ago

Q7: Determine the graph of the polar equation r = 12/6+4n cos theta.

iVinArrow [24]

Answer:

Choice B

Step-by-step explanation:

The first step is to write the polar equation of the conic section in standard form by dividing the numerator and denominator by 6;

$r=\frac{2}{1+\frac{2}{3}cos theta }$

The eccentricity of this conic section is thus 2/3, the coefficient of cos theta. Clearly, the eccentricity is between o and 1 implying that this conic section represents an Ellipse.

Lastly, the ellipse will open towards the left since we have positive cos theta in the denominator. The only graph that meets the conditions is graph B.

3 0

3 years ago

A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of

notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.01 = 0.99$ , so Z = 2.327.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031$

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.0001 = 2.327\frac{0.0003}{\sqrt{n}}$

$0.0001\sqrt{n} = 2.327*0.0003$

$\sqrt{n} = \frac{2.327*0.0003}{0.0001}$

$(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2$

$n = 48.73$

Rounding up

49 measurements are needed.

7 0

3 years ago

Find the population mean or sample mean as indicated. Sample: 19, 15, 6, 11, 24 Compute the sample mean for this data set. Selec

yKpoI14uk [10]

Answer:

$\overline{x}=15$

Step-by-step explanation:

the mean is given by:

$\overline{x} = \dfrac{\sum\limits_{i=1}^n x_i}{n} \quad\text{or}\quad \dfrac{\text{sum of all items}}{\text{number of items}}$

In our case this is:

$\overline{x} = \dfrac{19+15+6+11+24}{5} \Rightarrow \dfrac{75}{5}\\\\\overline{x} = 15\\\\$

side note: the main difference between sample mean and population mean is in the 'context'. However, the method to calculate them is the same.

By context I mean: if this the items are taken from some larger category for example: the ages of a few 'students' from a 'class'. Here 'students' are the sample from a larger set that is 'class'. The mean of the 'few students' will be called sample mean. In contrast, if we take the mean of the ages of the whole class then this is called population mean. (population mean == mean of the whole set)

In our case we aren't told exactly where these numbers come from, is this the whole set or a sample from it, the lack of context allows us to assume that the mean can either be population mean or sample mean. So we can safely use any symbol $\mu$ or $\overline{x}$ .

3 0

3 years ago

Find the 78th term of the arithmetic sequence<br> -12,-13,-14,

creativ13 [48]

Answer:

-89

Step-by-step explanation:

The first term of the sequence is a1 = -12.

The common difference between terms is ...

d = -13 -(-12) = -1

The generic term is given by ...

an = a1 +d(n -1)

Then the 78th term is ...

a78 = -12 +(-1)(78 -1) = -12 -77

a78 = -89

8 0

2 years ago