You can sort them on the shapes that are the same or different
Answer:
B. cos−1(StartFraction 11.9 Over 14.5 EndFraction) = θ
Step-by-step explanation:
From definition:
cos(θ) = adjacent/hypotenuse
The adjacent side respect angle GFE (or θ) is side FE, and side FG is the hypotenuse. Replacing with data and isolating θ:
cos(θ) = 11.9/14.5
θ = cos^-1(11.9/14.5)
<span>D = 180 * (n -2)
D = 180n -360
180n = D +360
n = (D / 180) + 2
So, the answer is "A"
And we can test this with a square, sum of angles = 360
n = (D / 180) +2
n = (360 / 180) +2
number of sides = 4
correct
Source:
http://www.1728.org/polygon.htm
</span>
Answer:
The sample consisting of 64 data values would give a greater precision.
Step-by-step explanation:
The width of a (1 - <em>α</em>)% confidence interval for population mean μ is:

So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (<em>n</em>).
That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.
Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.
The two sample sizes are:
<em>n</em>₁ = 25
<em>n</em>₂ = 64
The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.
Width for n = 25:
Width for n = 64:
![\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]](https://tex.z-dn.net/?f=%5Ctext%7BWidth%7D%3D2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7B64%7D%7D%3D%5Cfrac%7B1%7D%7B8%7D%5Ccdot%20%5B2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Csigma%5D)
Thus, the sample consisting of 64 data values would give a greater precision
Answer
given,
thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters.
X = U[0.95,1.05] 0.95≤ x ≤ 1.05
the cumulative distribution function of flange
F(x) = P{X≤ x}=
=
b) P(X>1.02)= 1 - P(X≤1.02)
= 
= 0.3
c) The thickness greater than 0.96 exceeded by 90% of the flanges.
d) mean = 
= 1
variance = 
= 0.000833