Can someone solve this? and tell me how?
1 answer:
Answer:
1/3
Step-by-step explanation:
When working with balanced expressions (stuff on both sides of the equal sign), "what you do to one side, you do to the other", which keeps it balanced.
The first thing we notice is the exponent 1/4, which is one both sides, so we can get rid of it on both sides by using the <u>reverse operation</u>.
The reverse of exponents is <u>square root</u>.
Isolate x to solve. Separate the variables and non-variables.
4x + 10 = 9 + 7x
4x - 4x + 10 = 9 + 7x - 4x Subtract 4x from both sides
10 = 9 + 3x
10 - 9 = 9 - 9 + 3x Subtract 9 from both sides
1 = 3x Divide both sides by 3 to isolate x
x = 1/3 Answer
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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
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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.
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