Since the Σ( of all colors )= 100%, OR 1, then:
a) P(GREEN ∪ BLUEU) = P(G) + P(BL) = 8%+6% = 14% or 0.14
Since we have to choose ONE candy and only ONE candy at random, then tey are mutually exclusive: No. Choosing a green and blue M&M is possible
b) P(YELLOW ∪ RED) = P(Y) + P(R) = 18%+18% = 36% or 0.36
SAME ANSWER AS BEFORE: mutually exclusive
c) P(NOT PURPLE), Let's calculate 1st, the probability of having a PURPLE:
P(PURPLE) = 21% or 0.21
And the Probability of NOT having a PURPLE is 1-0.21 = 0.79
A decagon has 10 sides.
It it is regular you can build 10 isosceles triangles from the center of the decagon to the 10 sides.
Each triangle has a common vertex where the angle of each triangle is 360° / 10 = 36°.
So each time that you rotate the decagon a multiple of 36° around the center you get an image that coincides with the original decagon.
If the letters are given clockwise:
- when you rotate 36° counter clockwise, the point A' (the image of A) will coincide with the point J.
- when you rotate 72° (2 times 36°) counter clockwise, the point A' will land on I.
- when you rotate 108° (3 times 36°) counter clockwise, the point A' will land on H.
- when you rotate 144° (4 times 36°) counter clockwise, the point A' will land on G.
- when you rotate 180° (5 times 36°) counter clockwise, the point A' will land on I.
- when you rotate 216° (6 times 36°) counter clockwise, the point A' will land on E.
- whn you rotate 252° (7 times 36°) counterclockwise, the point A' will coincide with D.
Add other 36° each time and A' will coincide successively with C, B and the same A.
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
Old
9.99×55
=549.45
New
10.68×55
=587.4
((10.68÷9.99)−1)×100
=6.9%