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2 years ago

How many times greater is the value of the digit 5 in 853,607 than the value of the digit 5 in 362,051.

Mathematics

1 answer:

Ray Of Light [21]2 years ago

4 0

Answer:

1,000 times greater

Step-by-step explanation:

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What is the volume of the figure below, which is composed of two cubes with side lengths of 7 units?

wariber [46]

Answer:

B) 686 cubic units

Step-by-step explanation:

If one cube has a side length of 7, then it equals to 343 cubic units. Since there are 2 cubes, this translates to 686 cubic units.

6 0

1 year ago

The low temperature on Monday was 6 degrees warmer than Sunday's low of -9°F.The low temperature on Tuesday was 3 degrees warmer

OverLord2011 [107]

Answer: the low temperature on Tuesday is 0 degree Fahrenheit

Step-by-step explanation:

The low temperature on Monday was 6 degrees warmer than Sunday's low of -9°F. This means that the temperature on Monday would be

- 9 + 6 = - 3 degrees Fahrenheit

The low temperature on Tuesday was 3 degrees warmer than Monday's low temperature. This means that the low temperature on Tuesday would be

- 3 + 3 = 0 degree Fahrenheit

5 0

2 years ago

Two identical square pyramids were joined at their bases to form the composite figure below. 2 square pyramids have a base of 24

Airida [17]

Answer:

$8[\frac{1}{2}(24)(5)]$ cm²

Step-by-step explanation:

Surface area of the composite figure = Surface area of the lateral surfaces of the given pyramids

Lateral surface area of a square pyramid = 4

Therefore, lateral surface area of the pyramid

= $4[\frac{1}{2}(24)(5)]$

Now lateral surface area of the composite figure = 2(lateral surface area of two pyramids)

= $8[\frac{1}{2}(24)(5)]$

Therefore, Expression for the surface area of the given composite figure is,

$8[\frac{1}{2}(24)(5)]$ cm²

6 0

2 years ago

Read 2 more answers

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]:

$\\ -0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}$

$\\ (-0.68 *0.21088) + 8.6 = \overline{X}$

$\\ \overline{x} =8.4566$

$\\ Q1 = 8.4566$ ppm.

For Q3

$\\ 0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}$

$\\ (0.68 *0.21088) + 8.6 = \overline{X}$

$\\ \overline{x} =8.7434$

$\\ Q3 = 8.7434$ ppm.

$\\ IQR = Q3-Q1 = 8.7434 - 8.4566 = 0.2868$ ppm

Therefore, the IQR for the average of 38 cities is $\\ IQR = 0.2868$ ppm. $\\ Q1 = 8.4566$ ppm and $\\ Q3 = 8.7434$ ppm.

4 0

2 years ago

What is the solution to this equation? 3[x] - 1 = 8

Yanka [14]

Answer:

Step-by-step explanation:

The whole goal is to isolate the variable.

3x-1=8

1+3x-1=8+1 first add 1 to both sides of the equation to cancel out the -1

3x=9 now just divide 3 to both sides of the equation because the opposite of multiplication is division. So dividing 3 on both sides cancels out the 3 to isolate the x

3x/3=9/3

x=3

8 0

2 years ago