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3 years ago

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The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population o

f x values has an approximately normal distribution.1229 1257 1313 1201 1268 1316 1275 1317 1275(a) Use a calculator with mean and standard deviation keys to find the sample mean year x and sample standard deviation s. (Round your answers to the nearest whole number.)x = A.D.s = yr(b) Find a 90% confidence interval for the mean of all tree ring dates from this archaeological site. (Round your answers to the nearest whole number.)lower limit A.D.upper limit A.D.

1 answer:

irga5000 [103]3 years ago

5 0

Answer:

1.The sample mean and standard deviation are 1272 and 40

2. The 90% confidence interval for the mean is between 1237 and 1306

Step-by-step explanation:

See attached picture for explanations.

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Simplify x 1/3(x1/2+2x²

tigry1 [53]

Answer:

$x^{\frac{5}{6} } +2x^\frac{7}{3}$

Step-by-step explanation:

1. x^1/3(x^1/2) = $x^{\frac{5}{6} }$

2. x^1/3(2x^2) = $2x^\frac{7}{3}$

3. = $x^{\frac{5}{6} }$ + $2x^\frac{7}{3}$

4 0

3 years ago

A random sample of 121 students from the University of Oklahoma had a sample mean ACT score of 23.4 with a sample standard devia

Grace [21]

Answer:

(22.74,24.06) is the required 95% confidence interval for the population mean ACT score of University of Oklahoma students.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 121

Sample mean = 23.4

Sample standard deviation = 3.65

Level of significance = 0.05

Degree of freedom

$= n - 1 = 120$

95% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Calculation of critical value:

$t_{critical}\text{ at degree of freedom 120 and}~\alpha_{0.05} = \pm 1.9799$

Putting the values, we get,

$23.4\pm 1.9799(\dfrac{3.65}{\sqrt{121}} )\\\\ = 23.4 \pm 0.6569\\\\ = (22.7431 ,24.0569)\approx (22.74,24.06)$

(22.74,24.06) is the required 95% confidence interval for the population mean ACT score of University of Oklahoma students.

5 0

3 years ago

Y = 5x - 10<br> Can you solve for x

tensa zangetsu [6.8K]

Answer:

$x = \frac{1}{5}y + 2$ or $x = \frac{y+10}{5}$ (Not sure which one is preferred in your case)

Step-by-step explanation:

<u>Key skills needed: Evaluating expressions</u>

1) We are given: $y = 5x-10$

2) To solve for the x variable, we want to leave the term with "x" by itself.

This means we add 10 to both sides

--> $y + 10 = 5x$ (Since -10 and +10 cancel out to make 0 or nothing)

3) Then we divide by 5 on both sides to get "x" completely by itself.

-----> $\frac{y+10}{5} = x$

4) You can keep it as is so --> $x = \frac{y + 10}{5}$

or you can divide "y" by 5 and "10" by 5 and get --> $x = \frac{1}{5} y + 2$

(I am not sure which form is preferred one is preferred as the teacher matters)

<em>Hope you understood and have a nice day!!</em>

6 0

3 years ago

Read 2 more answers

Zayn puts garland around a square window. He

mojhsa [17]

2 meters because 8/4 is 2

8 0

4 years ago

Given the Trapezoid ABCD<br><br> Find the length of AD <br><br> Pls hellpppp

matrenka [14]

Given:

In trapezoid $ABCD, AD\parallel BC, MN$ is the mid-segment of $ABCD, AD=30x-10,MN=31x+1,BC=30x+28$ .

To find:

The length of $AD$ .

Solution:

We know that the length of the mid-segment of a trapezoids is half of the sum of lengths of two parallel sides of the trapezoid.

In trapezoid $ABCD$ ,

$MN=\dfrac{AD+BC}{2}$

$31x+1=\dfrac{30x-10+30x+28}{2}$

$31x+1=\dfrac{60x+18}{2}$

$31x+1=30x+9$

Isolate variable terms.

$31x-30x=9-1$

$x=8$

Now,

$AD=30x-10$

$AD=30(8)-10$

$AD=240-10$

$AD=230$

Therefore, the length of $AD$ is 230.

6 0

3 years ago