There are two cases to consider.
A) The removed square is in an odd-numbered column (and row). In this case, the board is divided by that column and row into parts with an even number of columns, which can always be tiled by dominos, and the column the square is in, which has an even number of remaining squares that can also be tiled by dominos.
B) The removed square is in an even-numbered column (and row). In this case, the top row to the left of that column (including that column) can be tiled by dominos, as can the bottom row to the right of that column (including that column). The remaining untiled sections of the board have even numbers of rows, so can be tiled by dominos.
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Perhaps the shorter answer is that in an odd-sized board, the corner squares are the ones that there is one of in excess. Cutting out one that is of that color leaves an even number of squares, and equal numbers of each color. Such a board seems like it <em>ought</em> to be able to be tiled by dominos, but the above shows there is actually an algorithm for doing so.
Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean 
and standard deviation 
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:
X = 590:
Z = 0.76
Z = 0.76 has a p-value of 0.7764.
X = 400:
Z = -0.89
Z = -0.89 has a p-value of 0.1867.
0.7764 - 0.1867 = 0.5897 = 58.97%.
58.97% of students would be expected to score between 400 and 590.
More can be learned about the normal distribution at brainly.com/question/27643290
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Answer:
This is trapezium.
Area of this figure =1/2h(d1-d2)
1/2×5(14-8)
=30/2
=15cm^2
Answer: Are you ok?
Step-by-step explanation:
Answer:
35 percent of the 75,000 fans is: 0.35*75,000 = 26,250
since there are 2 fans per car we find the number of cars by dividing 26,250 by 2:
26,250 : 2 = 13,125 cars
5,000 will park at the stadium's parking lots, while the rest 13,125 - 5,000 = 8,125 cars will par at satellite lots.
OR
35% of 75000 = 26250
2 fans in a car
So, = 26250/2
= 13125
Total handling space = 5000
13125 - 5000 = 8125
Step-by-step explanation:
