What is gcf and lcm? idont understand
2 answers:
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Answer:
And the best option would be:
C. [17.42,20.78]
Step-by-step explanation:
Previous concepts
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".
represent the sample mean
population mean (variable of interest)
s=1.5 represent the sample standard deviation
n=9 represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
(1)
In order to calculate the critical value we need to find first the degrees of freedom, given by:
Since the Confidence is 0.99 or 99%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that
Now we have everything in order to replace into formula (1):
And the best option would be:
C. [17.42,20.78]
Answer:
Depth:
μ =20.2025 km
M = 15.625 km
Range = 47.15 km
σ ≈ 15.92 km
Q₁ = 5.7375 km
Q₃ = 34.6675 km
Magnitude:
μ = 2.08375
M = 1.465
Range, R = 5.17
σ = 1.801485 ≈ 1.8
Q₁ = 0.5625
Q₃ = 3.925
Step-by-step explanation:
The given data are;
Depth Magnitude
0.76 0.84
4.93 0.47
8.16 0.35
33.58 1.32
21.2 1.61
35.03 4.57
10.05 5.52
47.91 1.99
For the Depth, we have;
The mean, μ = (0.76+4.93+8.16+33.58+21.2+35.03+10.05+47.91)/8 =20.2025 km
The median, M = The (n + 1)/2th term after arranging the term in increasing order as follows;
0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 , the median is therefore;
(8 + 1)/2th term or the 4.5th term which is 10.05 + (21.2 - 10.05)/2 = 15.625 km
The Range = The highest - The lowest value = 47.91 - 0.76 = 47.15 km
The Standard deviation of, σ, is given as follows;
Where;
= The individual data point = (0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91
)
N = The total number of data point = 8
Substituting, (using Microsoft Excel) we get;
Q₁ = The first quartile = The (n + 1)/4th = term arranged in increasing order
Q₁ = The (8 + 1)/4th term = The 2.25th term = 4.93 + (8.16 - 4.93)×0.25) = 5.7375 km
Q₃ = The first quartile = The 3×(n + 1)/4th = term arranged in increasing order
Q₃ = The 3×(8 + 1)/4th term = The 6.75th term = 33.58 + 3×(35.03 - 33.58)×0.25) = 34.6675 km
For the magnitude, we have, using the same formulas and procedures as above;
μ = 2.08375
M = 1.465
Range, R = 5.17
σ = 1.801485 ≈ 1.8
Q₁ = 0.5625
Q₃ = 3.925