Twenty-five ten-thousandths in scientific notation?

Zina [86]

Hello,

$\dfrac{5}{7^3}=5*7^x\\
5*7^{-3}=5*7^x\\

\boxed{x=-3}
$

8 0

3 years ago

Suppose the method of tree ring dating gave the following dates A.D. for an archaeological excavation site. Please show your wor

natita [175]

Answer:

a) $\bar X=\frac{1245+1245+1321+1191+1295+1330+1239+1250+1228}{9}= 1260.444$

b) $s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s = 45.580$

c) For this case since the sampel size is n=9 <30 we can use the t distribution in order to find the confidence interval.

d) $1260.444-2.306 \frac{45.580}{\sqrt{9}}= 1225.408$

$1260.444+2.306 \frac{45.580}{\sqrt{9}}= 1295.480$

Step-by-step explanation:

For this case we ave the following data:

1245 1245 1321 1191 1295 1330 1239 1250 1228

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got:

$\bar X= \frac{1245+1245+1321+1191+1295+1330+1239+1250+1228}{9}= 1260.444$

Part b

For this case we can use the following formula in order to find the sample standard deviation:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s = 45.580$

Part c

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".

For this case since the sampel size is n=9 <30 we can use the t distribution in order to find the confidence interval.

Part d

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

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

We need to find the degrees of freedom given by:

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_{\alpha/2}=2.306$

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

$1260.444-2.306 \frac{45.580}{\sqrt{9}}= 1225.408$

$1260.444+2.306 \frac{45.580}{\sqrt{9}}= 1295.480$

8 0

4 years ago

The rectangular top of a small box has dimensions of one 40 mm x 230 mm. How long in centimeters with a ribbon need to go all th

kozerog [31]

Answer:

540 mm

Step-by-step explanation:

Here we are given a rectangular box with dimensions of the top surface as 40 mm and 230 mm

We are asked to determine the measurement of the ribbon which may go all the way around the edge of it. Basically we are being asked the perimeter of the top surface. The perimeter is given as the

P=2 (l+w)

l = 230

w = 40

P=2(230+40)

P=2 x 270

P= 540

Hence we need 540 mm of ribbon to go all the way around the edge of the top of the box.

8 0

3 years ago

Read 2 more answers

A $250 suede jacket is on sale for 20%off how much should you pay for the jacket

yarga [219]

250 * 0.20 = $50 sale
250 - 50 = $200
You will pay $200 after sale

4 0

3 years ago

Read 2 more answers

Qhat are the edge lengths of 6 inches and edge lenghthe of the other 1.25,4/3,1.5,7/4 and 2

finlep [7]

Given:U.S Flag - rectangle in shapescale of 0.5 cm = 1 inlength of the scale 2.5 cmwidth of the scale 4.75 cmperimeter of the scale = 2(L+W) = 2(2.5+4.75) = 2(7.25) = 14.5cmperimeter of the actual flag is:14.5cm * 1in/0.5cm = 29 inches.

Read more on Brainly.com - brainly.com/question/1306723#readmore

6 0

3 years ago