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Alex_Xolod [135]
3 years ago
15

Sam colors each tile in a 4 by 4 grid white or black. A coloring is called rotationally

Mathematics
1 answer:
DiKsa [7]3 years ago
8 0

Answer:

  65,280

Step-by-step explanation:

Consider the 4×4 grid ...

  \left[\begin{array}{cc}a&b\\d&c\end{array}\right]

where each of a, b, c, d is a 2×2 array of tiles. Let's use the notation a' to represent the 2×2 array "a" rotated right 1/4 turn. For 90° rotational symmetry, we must have b=a', c=b'=a'', d=c'=b''=a'''. That is, once "a" is determined, the rest of the grid is determined. Since "a" consists of 4 tiles, each of which can be black or white, there are 2^4 = 16 patterns that have 90° rotational symmetry.

The same will be true of 270° rotational symmetry, for the same reason.

__

For 180° rotational symmetry, we must have c=a'' and d=b''. Then the combination of "a" and "b" together fully determines the grid. Together, "a" and "b" consist of 8 tiles, so there are 2^8 = 256 ways to pattern the grid so it will have 180° rotational symmetry. (Of those, 16 have 90° symmetry, and 16 have 270° symmetry. The sets are overlapping.)

__

The 16 tiles of the grid can be colored 2^16 = 65,536 different ways. As we have seen, 256 of those colorings result in 180° rotational symmetry. Then the number of colorings that have no rotational symmetry is ...

  65,536 -256 = 65,280 . . . . colorings not rotationally symmetric

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Answer:

B. 196 units squared

Step-by-step explanation:

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I haven't did these type of questions in a while, but i think my answer is correct. Hope it helps :)

5 0
3 years ago
Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus
quester [9]
Hi! 

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20;  </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510

To calculate the degrees of freedom you need to use the following equation:

df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.



5 0
4 years ago
12.6 + 4m = 9.6 + 8m
Sergeu [11.5K]

Answer:

3 = 4m or  0.75 = m

Step-by-step explanation:

12.6 + 4m = 9.6 + 8m

Subtract 4m from both sides of the equation

12.6 = 9.6 + 4m

Subtract 9.6 from both sides of the equation

3 = 4m

Divide by four on both sides of the equation

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I Hope That This Helps! :)

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