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3 years ago

Write the expression in simplest form using only positive exponents (x^2/3y^-1/2)^-6

1 answer:

5 0

$\left( \cfrac{x^2}{3y^{- \frac{1}{2} }} \right)^{-6}=\left( \cfrac{3y^{- \frac{1}{2} }}{x^2} \right)^{6}=\left( \cfrac{3}{x^2y^{ \frac{1}{2} }} \right)^{6}= \cfrac{3^6}{x^{2*6}y^{ \frac{1}{2}*6 }} = \cfrac{729}{x^{12}y^3}$

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Please help me solve !!!!!!!!!!!!!!!

Dima020 [189]

Answer:

4x - 3y = 4

-3x + 3y = -6

x = -2

-2 - y = 2

-y = 4

y = -4

(-2,-4)

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3 years ago

to determine a student's grade, a teacher throws out the lowest grade obtained on 5 tests, averages the remaining grades, and ro

RoseWind [281]

Answer:

B grade

Step-by-step explanation:

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3 0

3 years ago

Read 2 more answers

Suppose the random variables X, Y, and Z have the following joint probability distribution. x y z f ( x , y , z ) 1 1 1 0.05 1 1

Airida [17]

Answer:

Determine the conditional probability distribution of X given that Y = 1 and Z = 2. Round your answers to two decimal places (e.g. 98.76).

answer:

Given that Y = 1 : 2/5

Given that Z = 2 : 3/5

Step-by-step explanation:

The conditional probability distribution of X F x | yz^( x )

Given that Y = 1

F x | yz . ( x | yz ) = 2/5

Given that z = 2

= 3/5

attached below is the detailed solution

3 0

3 years ago

Find the measure of line segment DF. A) 12 B) 13 C) 14 D) 15

brilliants [131]

As per the Intersecting Chords theorem, if two chords cross in a circle, the products of the chord segments' measures are equal. The measure of the line segment is DF is 14 units.

<h3>What is Intersecting Chords theorem?</h3>

If two chords cross in a circle, the products of the chord segments' measures are equal.

As per the Intersecting Chords theorem, the product of DF and FB will be equal to the product of CF and FA, therefore,

DF×FB = CF×FA

(x+8) × 8 = 16 × 7

(x+8) = (16×7)/8

(x+8) = 14

Hence, the measure of the line segment is DF is 14 units.

Learn more about Intersecting Chords theorem:

brainly.com/question/15298662

#SPJ1

4 0

2 years ago

The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.6 ppm and standard de

Setler79 [48]

We assume that question b is asking for the distribution of $\\ \overline{x}$ , that is, the distribution for the average amount of pollutants.

Answer:

a. The distribution of X is a normal distribution $\\ X \sim N(8.6, 1.3)$ .

b. The distribution for the average amount of pollutants is $\\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}})$ .

c. $\\ P(z>-0.08) = 0.5319$ .

d. $\\ P(z>-0.47) = 0.6808$ .

e. We do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for $\\ \overline{X}$ is also normal because the sample was taken from a normal distribution.

f. $\\ IQR = 0.2868$ ppm. $\\ Q1 = 8.4566$ ppm and $\\ Q3 = 8.7434$ ppm.

Step-by-step explanation:

First, we have all this information from the question:

The random variable here, X, is the number of pollutants that are found in waterways near large cities.
This variable is normally distributed, with parameters:
$\\ \mu = 8.6$ ppm.
$\\ \sigma = 1.3$ ppm.
There is a sample of size, $\\ n = 38$ taken from this normal distribution.

a. What is the distribution of X?

The distribution of X is the normal (or Gaussian) distribution. X (uppercase) is the random variable, and follows a normal distribution with $\\ \mu = 8.6$ ppm and $\\ \sigma =1.3$ ppm or $\\ X \sim N(8.6, 1.3)$ .

b. What is the distribution of $\\ \overline{x}$ ?

The distribution for $\\ \overline{x}$ is $\\ N(\mu, \frac{\sigma}{\sqrt{n}})$ , i.e., the distribution for the sampling distribution of the means follows a normal distribution:

$\\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}})$ .

c. What is the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants?

Notice that the question is asking for the random variable X (and not $\\ \overline{x}$ ). Then, we can use a standardized value or z-score so that we can consult the standard normal table.

$\\ z = \frac{x - \mu}{\sigma}$ [1]

x = 8.5 ppm and the question is about $\\ P(x>8.5)$ =?

Using [1]

$\\ z = \frac{8.5 - 8.6}{1.3}$

$\\ z = \frac{-0.1}{1.3}$

$\\ z = -0.07692 \approx -0.08$ (standard normal table has entries for two decimals places for z).

For $\\ z = -0.08$ , is $\\ P(z$ .

But, we are asked for $\\ P(z>-0.08) \approx P(x>8.5)$ .

$\\ P(z-0.08) = 1$

$\\ P(z>-0.08) = 1 - P(z$

$\\ P(z>-0.08) = 0.5319$

Thus, "the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants" is $\\ P(z>-0.08) = 0.5319$ .

d. For the 38 cities, find the probability that the average amount of pollutants is more than 8.5 ppm.

Or $\\ P(\overline{x} > 8.5)$ ppm?

This random variable follows a standardized random variable normally distributed, i.e. $\\ Z \sim N(0, 1)$ :

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

$\\ z = \frac{\overline{8.5} - 8.6}{\frac{1.3}{\sqrt{38}}}$

$\\ z = \frac{-0.1}{0.21088}$

$\\ z = \frac{-0.1}{0.21088} \approx -0.47420 \approx -0.47$

$\\ P(z$

Again, we are asked for $\\ P(z>-0.47)$ , then

$\\ P(z>-0.47) = 1 - P(z$

$\\ P(z>-0.47) = 1 - 0.3192$

$\\ P(z>-0.47) = 0.6808$

Then, the probability that the average amount of pollutants is more than 8.5 ppm for the 38 cities is $\\ P(z>-0.47) = 0.6808$ .

e. For part d), is the assumption that the distribution is normal necessary?

For this question, we do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for $\\ \overline{X}$ is also normal because the sample was taken from a normal distribution. Additionally, the sample size is large enough to show a bell-shaped distribution.

f. Find the IQR for the average of 38 cities.

We must find the first quartile (25th percentile), and the third quartile (75th percentile). For $\\ P(z$ , $\\ z \approx -0.68$ , then, using [2]: