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3 years ago

In this diagram, each square represents 1 km. What is the displacement of the car?

Mathematics

2 answers:

Brut [27]3 years ago

8 0

<h3>Answer: C) 2 km west</h3>

Explanation:

With displacement, all we care about is the beginning and end. We don't care about the middle part(s) of the journey. So we'll take the straight line route from beginning to end when it comes to computing displacement.

We start at A(0,0) and end at B(-2,0). Going from A directly to B has us go 2 km west. Keep in mind that displacement is a vector, so you must include the direction along with the distance.

morpeh [17]3 years ago

6 0

Answer:

Step-by-step explanation:

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Given angle DOG, what is the third step with you construct the copy of angle DOG?

Leto [7]

The correct answer is choice C

7 0

3 years ago

Solve the equation<br> 2x-3=15<br> A)x=6<br> B)x=9<br> C)x=24<br> D)x=36

DaniilM [7]

To solve the equation, it is necessary to isolate the unknown variable "x" first. This is done as shown below:

2x - 3 = 15

Adding 3 to both sides of the equation:

2x - 3 + 3 = 15 + 3
2x = 18

Dividing 2 from both sides:

2x/2 = 18/2
x = 9

Among the choices, the correct answer is B.

5 0

4 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 $t_{\alpha/2}=2.306$