Which expression is equivalent to (a^-3 b/a^-5 b3)^-3? Assume a=0, b=0.

1 answer:

7 0

$Use:\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\\\(ab)^n=a^nb^n$

$\left(\dfrac{a^{-3}b}{a^{-5}b^3}\right)^{-3}=\left(a^{-3-(-5)}b^{1-3}\right)^{-3}=\left(a^{-3+5}b^{-2}\right)^{-3}=\left(a^2b^{-2}\right)^{-3}\\\\=(a^2)^{-3}(b^{-2})^{-3}=a^{2(-3)}b^{-2(-3)}=a^{-6}b^6=\dfrac{b^6}{a^6}$

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Simplify the expression p(p^-7*q^3)^-2*q^-3<br> A.p/q^3<br> B.p^15/q^9<br> C.1/(p^9*q^2<br> D.p^14*q

Alenkinab [10]

Answer:

B.p^15/q^9

Step-by-step explanation:

p(p^-7•q^3)^-2•q^-3

answer: B. p^15/q^9

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3 years ago

Majesty Video Production Inc. wants the mean length of its advertisements to be 26 seconds. Assume the distribution of ad length

Paladinen [302]

Answer:

a) By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.

b) s = 0.44

c) 0.84% of the sample means will be greater than 27.05 seconds

d) 98.46% of the sample means will be greater than 25.05 seconds

e) 97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds

Step-by-step explanation:

To solve this question, we need 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 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(also called standard error) $s = \frac{\sigma}{\sqrt{n}}$ .