Zucchini weights are approximately normally distributed with mean 08 pound and standard deviation 0.25 pound. Which of the follo
wing shaded regions best represents the probability that a
randomly selected zucchini will weigh between 0 55 pound and 13 pounds?
randomly selected zucchini will weigh between 0 55 pound and 13 pounds?
1 answer:
Answer:
The correct option is A
Step-by-step explanation:
From the question we are told that
The mean is
The standard deviation is
Generally the the probability that a randomly selected zucchini will weigh between 0 55 pound and 13 pounds is mathematically represented as
=>
So looking the option of normal curves given in the question we see that it is option A that matches our interval
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we know that t is the number of years that plant is alive.
we know that is expression representing the number of bees that will visit the plant in its life time.
Answer:
- a)
- b)
- c)
- d)
Step-by-step explanation:
We will use the product rule from combinatorics.
- a) There are 26 letters in the English alphabet, so there are 26 possible choices for the first character and 26 possible choices for the last one. Each one of the remaining eight characters of the string has 36 choices (letters or digits). By the product rule, there are
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- b) We have 5 possible choices for the first character, it must be some vowel a,e,i,o,u. The second character can be chosen in 21 ways, selecting some consonant. There are 10 possibilities for the last character because only of the digits are allowed. The other seven characters have no restrictions, so each one can be chosen in 36 ways. By the product rule there are
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- c) The third character has 5 possibilities. Repetition of vowels is allowed, so the sixth and eighth characters have each one 5 possible choices. There are seven characters left. None of them are a vowel, but they are allowed to take any other letter or digit, so each one of them can be chosen in 36-5=31 ways. Therefore there are
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- d) Remember that the binomial coefficient
is the number of ways of choosing k elements from a set of n elements. In this case, to count all the possible strings, we first need to count in how many ways we can select the four positions that will have the digits. This can be done in
ways, since we are choosing four elements from the set of the ten positions of the string. Now, for the first position, we can choose any digit so it has 10 possibilities. The second position has 9 possibilities, because we can't repeat the digit used on the first position. Similarly, there are 8 choices for the third position and there are 7 choices for the fourth. Now, these are the only digits on the string, so the remaining 6 characters must be letters, then each one of them has 26 possibilities. By the product rule, there are
strings.