How to write 15,409 in word form

(01.02 LC)

ANTONII [103]

<h2>Answer:</h2><h2>x = -15</h2><h2>Hope this helps!!</h2>

7 0

3 years ago

B) The number of volume of a cylinder is half of its number of the total surface area.

faust18 [17]

Answer:

$V=179.6cm^3$

Step-by-step explanation:

the volume of a cylinder is given by:

$V=\pi r^2h$

where r is the radius and h is the height.

and the surface area:

$SA=2\pi r^2+2\pi rh$

the first term is the area of the circles and the second term is the area of the body.

since "The number of volume of a cylinder is half of its number of the total surface area." we will have that:

$V=\frac{1}{2}SA$

substitutig the equivalent expressions on each side:

$\pi r^2 h = \frac{1}{2} (2\pi r^2+2\pi r h)$

and we simplify and solve for the height (since is the value we don't know of the cylinder):

$\pi r^2 h = \pi r^2+\pi r h\\\pi r^2h-\pi rh=\pi r^2\\h(\pi r^2-\pi r)=\pi r^2\\h=\pi r^2/(\pi r^2-\pi r)$

we substitute the value of the radius $r=7cm$ , and we get:

$h=\pi (7cm)^2/(\pi (7cm)^2-\pi (7cm))\\h=153.938/(153.938-21.991)\\h=153.938/131.947\\h=1.1666cm$

thus the volume is:

$V=\pi r^2h$

$V=\pi (7cm)^2(1.1666cm)\\V=179.6cm^3$

3 0

2 years ago

Read 2 more answers

An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th

monitta

Answer:

a) n =1623

b) $ME=2.0538\sqrt{\frac{0.21 (1-0.21)}{1623}}=0.0208$

Step-by-step explanation:

1) Notation and definitions

$n$ random sample taken

$\hat p=0.19$ estimated proportion of customers are not satisfied with the service provided by the local dealer

Confidence =0.96 or 96%

Me= 0.02 represent the margin of error

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 96% of confidence, our significance level would be given by $\alpha=1-0.96=0.04$ and $\alpha/2 =0.02$ . And the critical value would be given by:

$z_{\alpha/2}=-2.0538, z_{1-\alpha/2}=2.0538$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.02$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.19(1-0.19)}{(\frac{0.02}{2.0538})^2}=1622.912$

And rounded up we have that n=1623

Part b

The margin of error is given by:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

So then if w replace the value of n obtained from part a we got:

$ME=2.0538\sqrt{\frac{0.21 (1-0.21)}{1623}}=0.0208$

3 0

3 years ago

You buy shoes for $57.84. the sales tax rate is 4%. estimate the sales tax. show your work.

aivan3 [116]

4% of $57.84
= 2.31 est. is sales tax

3 0

3 years ago

A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec

VMariaS [17]

Answer:

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

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".

$\bar X=19.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 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:

$df=n-1=9-1=8$

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(0.005,8)".And we see that $t_{\alpha/2}=$

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

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

5 0

3 years ago