$\text{Use}\ a^{-n}=\dfrac{1}{a^n}\\\\\dfrac{5^{-2}\cdot9}{2^4\cdot4^{-3}}=(5^{-2}\cdot9):(2^4\cdot4^{-3})=\left(\dfrac{1}{5^2}\cdot9\right):\left(2^4\cdot(2^2)^{-3}\right)\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=\left(\dfrac{1}{25}\cdot9\right):(2^4\cdot2^{-6})\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\dfrac{9}{25}:2^{4+(-6)}=\dfrac{9}{25}:2^{-2}=\dfrac{9}{25}:\dfrac{1}{2^2}=\dfrac{9}{25}:\dfrac{1}{4}=\dfrac{9}{25}\cdot\dfrac{4}{1}=\dfrac{36}{25}$

7 0

3 years ago

How do you solve this problem

torisob [31]

Divide the price by the number of payments.

$15,480/36 = $430

4 0

2 years ago

Examination of court records in a particular state shows that the mean sentence length for first-offense drug dealers is 26 mont

Zigmanuir [339]

Answer and explanation:

Given : Examination of court records in a particular state shows that the mean sentence length for first-offense drug dealers is 26 months with a standard deviation of 2 months. The records show that the sentence lengths are normally distributed.

i.e. Mean $\mu=26$ , Standard deviation $\sigma=2$

Using Z-score formula,

$Z=\frac{\bar{x}-\mu}{\sigma}$

1) What is the Z score for a 23 month sentence length? What is the probability of getting a sentence below that?

The Z score value for a $\bar{x}=23$ is given by,

$Z=\frac{23-26}{2}$

$Z=\frac{-3}{2}$

$Z=-1.5$

The probability of getting a sentence below that is $P(X$

i.e. $P(X$

$P(X$

2) A defense attorney is concerned that his client's sentence was unusually harsh at 30 months. What percent of sentences are 30 months or longer? Calculate the Z score and report the area.

The Z score value for a $\bar{x}=30$ is given by,