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1 year ago

Does anybody know this? Evaluate the following expression using the correct order of operations.

Mathematics

1 answer:

krok68 [10]1 year ago

6 0

Answer:

-4

Step-by-step explanation:

Wehn dealing with order of operations, we can remember the expression PEMDAS, which stands for parentheses, exponents, multiplication, division, addition, and subtraction.

So, our first step in the above problem is the parentheses:

$2(3-5)^3+4(-4+7)\\2(-2)^3+4(3)$

The second step is exponents:

$2(-2)^3+4(3)\\2(-8)+4(3)$

The third step is multiplication:

$2(-8)+4(3)\\-16+12$

The fourth and final step is addition:

$-16+12\\-4$

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3 years ago

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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:

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

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

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

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

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3 years ago

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Answer:

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Step-by-step explanation:

Before finding the test statistic, we need to understand the central limit theorem and subtraction of normal variables.

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

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