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3 years ago

3x^2 - 4x + 9x - x^2

Mathematics

2 answers:

fenix001 [56]3 years ago

5 0

Answer:

2x^2 +5x

Step-by-step explanation:

3x^2 - 4x + 9x - x^2

Combine like terms

3x^2 -x^2 -4x+9x

2x^2 +5x

Ray Of Light [21]3 years ago

3 0

Answer:

Just combine like terms:

2x^2+5x

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iris [78.8K]

Its easy you just have to do the pi formula

3 0

3 years ago

Find x and y so that the quadrilateral is a parallelogram.

sergiy2304 [10]

Answer:

Result:

The value of x = 12

The value of y = 21

Step-by-step explanation:

Given

The parallelogram DEFG

DE = 6x-12

FG = 2x+36

EF = 4y

DG = 6y-42

We know that the opposite sides of a parallelogram are equal.

As DE and FG are opposite sides, so

DE = FG

substituting DE = 6x-12 and FG = 2x+36 in the equation

6x-12 = 2x+36

6x-2x = 36+12

simplifying

4x = 48

dividing both sides by 4

4x/4 = 48/4

x = 12

Therefore,

The value of x = 12

Also, EF and DG are opposite sides, so

EF = DG

substituting EF = 4y and DG = 6y-42 in the equation

4y = 6y-42

switching sides

6y-42 = 4y

6y-4y = 42

2y = 42

dividing both sides by 2

2y/2 = 42/2

y = 21

Therefore,

The value of y = 21

Result:

The value of x = 12

The value of y = 21

3 0

3 years ago

Find the sum and express it in simplest form.

aksik [14]

Answer:

4x-2

Step-by-step explanation:

8 0

3 years ago

Penny Weights, use the weights of the post 1983 pennies to construct a 98% confidence interval estimate of the standard deviatio

Ronch [10]

Answer:

c. 0.1391 < Ï < 0.2311

Step-by-step explanation:

The formula for Confidence Interval is given as:

Mean ± z × Standard deviation/√n

Z score for 98% confidence interval = 2.326

Mean = Significance level = 100% - 98%

= 2% = 0.02

Standard deviation = S=0.0165

n= 37

Hence,

Confidence Interval =

0.02 ± 2.326 × 0.0165/√37

0.02 ± 0.0063094687

Confidence Interval

0.02 - 0.0063094687

= 0.1391

0.02 ± 0.0063094687

= 0.2311

Hence, the Confidence Interval = (0.1391, 0.2311)

= c. 0.1391 < Ï < 0.2311

7 0

3 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]: