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3 years ago

Write the multiplication equation that the diagram represents

Mathematics

1 answer:

Maurinko [17]3 years ago

6 0

1/2 times 1/2 is your answer, because all the diagrams represent that

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Which is the graph of f(x) = -3_/x?

Delicious77 [7]

Answer:

Step-by-step explanation:

sorry if it wrong

6 0

3 years ago

X squared minus 6x equals 0

Masja [62]

x²-6x=0

x(x-6)=0

Two solutions are possible:

x=0

x=6

8 0

3 years ago

Figure A

Amiraneli [1.4K]

<h2>Answer:</h2>

Figure B

<h2>Step-by-step explanation:</h2>

The Pythagorean Theorem is $a^2 + b^2 = c^2$ , where c is the longest side of the triangle (the hypotenuse).

To find the side length of each square, you have to square root the area of each square. This means that Figure A has side lengths of 3, 6 and 8 units. Figure B has side lengths of 5, 12 and 13 units.

In Figure A, if the triangle is right-angled, the equation $3^2 + 6^2 = 8^2$ must be correct. 9 + 36 = 45. 45 is not equal to 64, so the triangle is not right-angled.

In Figure B, if the triangle is right-angled, the equation $5^2 + 12 ^2 = 13^2$ must be correct. 25 + 144 = 169. 169 is 13 squared, so the triangle is right-angled.

Alternatively, as you are already given the square values for each side length, there is no need to square root and square again. You can just test if the two smaller areas equal the larger area, but the explanation above uses a more detailed example of the Pythagorean Theorem.

8 0

2 years ago

Read 2 more answers

What is the volume of the given cone?

faust18 [17]

Answer:

we have

radius[r]=9ft

height [h]=8ft

now

Volume of cone =1/3×πr²h=1/3×π×9²×8=678.584ft³

5 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 <em>the sample was taken from a normal distribution</em>.

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 <em>normally distributed</em>, 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 <em>standardized value</em> or <em>z-score</em> so that we can consult the <em>standard normal table</em>.

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