Can anyone help please?

Simplify the following expression as a monomial<br><br> X^2y÷yx^2

Novay_Z [31]

$\bf ~\hspace{7em}\textit{negative exponents}
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a^{-n} \implies \cfrac{1}{a^n}
~\hspace{4.5em}
a^n\implies \cfrac{1}{a^{-n}}
~\hspace{4.5em}
\cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m}
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\rule{34em}{0.25pt}\\\\
x^2\div yx^2\implies \cfrac{x^2}{yx^2}\implies \cfrac{x^2\cdot x^{-2}}{y}\implies \cfrac{x^{2-2}}{y}\implies \cfrac{x^0}{y}\implies \cfrac{1}{y}$

8 0

3 years ago

Kinda confused on what the sum would be?!? HELP NOWW PLEASE

valentinak56 [21]

That is b. aka 2
thanks

4 0

3 years ago

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .