What are four consecutive odd integers with the sum of 232

1 answer:

7 0

Four consecutive odd integers, eh? Let x be the first of these odd integers. The next odd integer is thus x+2, and the next is x+4, and the last is x+6. The sum of all four of these is 232, so you have x+x+2+x+4+x+6=2324x+12=232

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What is 3,4,5 plz and thank you

Tcecarenko [31]

For 3 it’s b. For 2 it’s 8. For 4 it’s the whole number

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3 years ago

You take a quiz with 6 multiple choice questions. After you studied your estimated that you would have about an 80 % chance of g

Arada [10]

Answer:

1.15%

Step-by-step explanation:

To get the probability of m independent events you multiply the individual probability of each event. In this case we have m independent events, each one with the same probability, therefore:

$p^{m}$

$0.8^{20} = 1.15\%$

This is a particlar scenario of binomial distribution problem. So the binomial distribution questions are about the number of success of m independent events, where every individual event has the same p probability. In the question we have 20 events and each event has a probability of 80%. The binomial distribution formula is:

$\binom{n}{k} * p^{k} * (1-p)^{n-k}$

n is the number of events

k is the number of success

p is the probability of each individual event

$\binom{n}{k}$ is the binomial coefficient

the binomial coefficient allows to find the subsets of k elements in a set of n elements. In this case there is only one subset possible since the only way to get 20 of 20 correct questions is to getting right all questions (for getting 19 of 20 questions there are many ways, for example getting the first question wrong and all the other questions right, or getting second questions wrong and all the other questions right, etc).

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$