You are absolutely correct. :)
Btw, here's my take on the christmas tree problem...
Let the blue, red and white balls be labelled A, B, C, respectively. Weigh A1+B1 against A2+C1. If A1+B1 == A2+C1, weigh A1 versus A2. If A1 > A2, then the heavier balls are A1, B2 and C1; otherwise, it is...
I think I am able to get it in 4 weighings (but not less). The initial steps are as follows:
Assuming that the coins are labelled 1 to 8, then the initial weighing is 12vs34. If 12=34, then the second weighing is 156vs278. Otherwise (whether 12>34 or 12<34), the second weighing is 137vs248...
Oops, it should have been:
Remove the path (say of length x) from the garden, we have the area equation
(55)(40) - x = \sqrt{x^2 - 40^2} (40)
and that gives x = 66.67 yards.
Answer is hidden.
Remove the path (say of length x) from the garden, we have the area equation
(55)(40) - x = 2 \sqrt{x^2 - 40^2} (40)
and that gives x = 48.202 yards.
Happy New Year to you too; yes, after much experimentation with 'staircase' cuts and with the initial assumption that the 8x1 is located at the center of the room.
Solution hidden below. As a hint, the 8x1 is placed exactly vertically at the center of the room. :)
1. Cut the 10x10 into 2 portions.
AAAAAAAAAB
AAAAAAAAAB
AAAABAAABB
AAAABAAABB
AAABBAABBB
AAABBAABBB
AABBBABBBB
AABBBABBBB
ABBBBBBBBB
ABBBBBBBBB
2. For portion A, move the whole...
Answer is hidden.
The probability of getting n heads and (N-n) tails in N toss is
P(n) = \binom{N}{n} (0.5)^N
and the corresponding multiplier of gamble is
X(n) = (1.5)^n (0.6)^{N-n} = (0.6)^N (2.5)^n
The mean of this measure is
E{X} = \sum_{n=0}^N P(n) X(n) = (0.3)^N \sum_{n=0}^N...
You could do this...
xbuff = [];
for k = 1:(the number of tones)
x = (insert your function here);
xbuff = [xbuff; x(:)];
end
sound(xbuff,fs);
wavwrite(xbuff,fs,bits,'audio.wav');
Express the convolution as an integral. Compare this integral with the problem statement and obtain a function of the form h(t-\lambda). It should not be too difficult then to determine h(t).