1. (10 pts.) Suppose you have a language with only 3 letters: A, B, and C, which occur with frequencies 0.7, 0.2, and 0.1, respe
ctively. The following ciphertext was encrypted by the Vigen`ere method: BABABCBCAC. Assume that the key length is either 1, 2, or 3. Find the most likely key length and determine the most likely key using the method from class. Solution: We compare BABABCBCAC with its shifts by 1, 2, and 3, looking for overlaps. There are no overlaps for a shift of 1 or 3, but there are 6 overlaps for a shift of 2. Thus we expect the key length to be 2. Looking at every other letter, we see BBBBA and AACCC for the letters in even and odd positions, respectively. Due to the overwhelming frequency of the letter A in the language, we can reasonably assume that B decrypts to to A in the first sequence and C decrypts to A in the second. This gives us the key BC.
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Answer:
50 kg water.
Step-by-step explanation:
We have been given that the number of kilograms of water in a human body varies directly as the mass of the body.
We know that two directly proportional quantities are in form , where y varies directly with x and k is constant of variation.
We are told that an 87-kg person contains 58 kg of water. We can represent this information in an equation as:
Let us find the constant of variation as:
The equation represents the relation between water (y) in a human body with respect to mass of the body (x).
To find the amount of water in a 75-kg person, we will substitute in our given equation and solve for y.
Therefore, there are 50 kg of water in a 75-kg person.