Angle I measured 20 degrees, side HI = 5, angle L = 20 degrees, and side KL =5. What additional information would you need to pr
ove that triangle HIJ is congruent to triangle KLM by SAS
2 answers:
Answer:
IJ = LM.
Step-by-step explanation:
The Side-angle-side postulate states that if the pair of a corresponding sides and angles formed between these sides of two triangles are equal in measurement then these two triangle are said to be congruent.
As shown in figure 1 below:
In Δ HIJ and ΔKLM,
∠I = ∠L = 20° and HI = KL = 5 units
Since the ∠I lies between HI and IJ and ∠L lies between KL and LM, ∴ if IJ = LM then Δ HIJ and ΔKLM will be congruent by SAS postulate.
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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.
For more data please see the image below.
#SPJ4
Answer:
Step-by-step explanation:
-This is an LCM problem.
-To simplify, we introduce a least common multiplier which is equivalent the product of the denominators:
#We introduce the LCM and adjust the fractions based on it :
Hence, the simplified form of the fraction is: