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raketka [301]
3 years ago
14

Nth term of 8,11,16,23

Mathematics
1 answer:
Dmitry [639]3 years ago
3 0
The ninth term will be 71.
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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
blondinia [14]

The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

According to the statement

we have to find that the standard deviation, mean and the intervals from the given data.

So, According to the given data from the method of tree ring dating

The value of mean is

x-bar = (1271 + 1208 + 1229 + 1299 + 1268 + 1316 + 1275 + 1317 + 1275) / 9 = x-bar = 1273 AD

And now we find standard deviation :

s = √∑(xi - x-bar) / (N - 1)

∑(xi - x-bar)^2 = (1271 - 1273)2 + (1208 - 1273)2 + (1229 - 1273)2 + ... + (1275 - 1273)2

∑(xi - x-bar)^2 = (-2)2 + (-65)2 + (-44)2 + ... + (2)2

∑(xi - x-bar)^2 = 4 + 4225 + 1936 + 676 + 25 + 1849 + 4 1936 + 4

∑(xi - x-bar)^2 = 10,659

Now,

s^2 = 10659/8 = 1332

s = 37 years

So, standard deviation is 37 years.

We need the t-distribution table since the standard deviation is unknown.  Therefore, our degrees of freedom is 9 - 1 = 8 and the critical value is 1.860.  Set up the confidence interval for the mean:

[x-bar ± t*(s/√n)] = [1273 ± 1.860*(37/√9)]

[x-bar ± t*(s/√n)] = [1250,1296]

So, The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

Learn more about method of tree ring dating here

brainly.com/question/15107034

Disclaimer: This question was incomplete. Please find the full content below.

Question:

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.

For more data please see the image below.

#SPJ4

4 0
1 year ago
The distance from Earth to Mars is 136,000,000 miles. A spaceship travels at 31,000 miles per
posledela
136,000,000/31,000= how many hours for the spaceship to reach mars
136,000,000/31,000/24= how many days for the spaceship to reach mars
136,000,000/31,000/24 is approx. 182.7 days or 183 days (rounded up b/c of "to the nearest day").
7 0
3 years ago
Read 2 more answers
How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%
blondinia [14]

The trick is to exploit the difference of squares formula,

a^2-b^2=(a-b)(a+b)

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2

Whatever you do to the denominator, you have to do to the numerator too. So

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2

Expand the numerator:

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}

So we have

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2

But √12 = √(3•4) = 2√3, so

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}

7 0
3 years ago
Ax + 4y = 5z, for a​
Stolb23 [73]

Answer:A=5z-4y/x

Step-by-step explanation:

You must bring 4y over first making it negative then to separate the x from the A you have to divide

4 0
3 years ago
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Is seven eighths greater than two eights
o-na [289]

It is because there are 7 1/8 and it is bigger than 2 1/8.
8 0
3 years ago
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