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3 years ago

Evaluate the following expression (-3)^-2

Mathematics

2 answers:

OleMash [197]3 years ago

8 0

<h3>Answer: the fraction 1/9</h3>

It is approximately equal to 0.1111

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To get this answer, we use the rule

$x^{-k} = \frac{1}{x^k}$

The negative exponent tells us to apply the reciprocal to get the exponent to be positive.

So,

$(-3)^{-2} = \frac{1}{(-3)^2}\\\\(-3)^{-2} = \frac{1}{9}$

Squaring -3 means you square the negative as well

$(-3)^2 = (-3)*(-3) = 9$

svlad2 [7]3 years ago

7 0

Answer:

1/9

Step-by-step explanation:

(-3)^-2 would be better written as (-3)^(-2).

(-3)^(-2) would be easier to evaluate if written as -------------

(-3)^2

The final answer is 1/9.

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3 years ago

The mean weight of boxes shipped by a company is 12 lbs., with a standard deviation of 4 lbs. Assume weight of boxes follows Nor

777dan777 [17]

Answer:

0.89% probability that a palette of 10 boxes will exceed that limit (150 lbs.)

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 12, \sigma = 4, n = 10, s = \frac{4}{\sqrt{10}} = 1.2649$

1. What’s the probability that a palette of 10 boxes will exceed that limit (150 lbs.)? a. HINT: Prob (total weight of a sample of 10 boxes >150 lbs.) = Prob (the sample mean weight ( y hat ) > 15 lbs.)

This is 1 subtracted by the pvalue of Z when X = 15. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$