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Taya2010 [7]
3 years ago
13

Elaina wrote two correct variable expressions to represent the phrase “one-fourth of a number.” Which variable expressions could

Elaina have written? Select each correct answer.
A.) 4 ÷ n
B.) 4/n
C.) 1/4n
D.) n4
Mathematics
2 answers:
Naily [24]3 years ago
7 0
1/4n, and also D IF it's n/4.
Vesna [10]3 years ago
3 0

Answer:

\frac{1}{4}n

\frac{n}{4}

C and D are correct options.

Step-by-step explanation:

The given phrase is “one-fourth of a number.”

Let the number be n.

one fourth means 1/4

Hence, the phrase “one-fourth of a number.” can be written as

\frac{1}{4}n

We can rewrite this expression by multiplying numerator as

\frac{n}{4}

Therefore, the required expressions are

\frac{1}{4}n

\frac{n}{4}

C and D are correct options.

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12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests
Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem, we have that:

12% of children are nearsighted. This means that p = 0.12.

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so n = 18

This probability is:

P = P(X = 0) + P(X = 1)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{18,0}.(0.12)^{0}.(0.88)^{18} = 0.1002

P(X = 1) = C_{18,1}.(0.12)^{1}.(0.88)^{17} = 0.2458

So

P = P(X = 0) + P(X = 1) = 0.1002 + 0.2458 = 0.3460

There is a 34.60% probability that 0 or 1 of them is nearsighted.

5 0
3 years ago
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