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2 years ago

Can someone please help me with this math problem. It is rate of change.

Mathematics

1 answer:

Shalnov [3]2 years ago

8 0

Answer:

Step-by-step explanation:

I think 2700miles /(4+2/3) equal 578,5

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3 years ago

Market research shows that 30% of consumers are likely to purchase an electronic tablet for younger children under age 10. Howev

Yuliya22 [10]

The probability of both children receiving a tablet is 15%.

We have given that

Age of Children P(Tablet) P(Laptop)

Less than 10 0.3 0.02

10-17 0.5 0.4

<h3>What is the probability?</h3>

Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.

Probability of both children receiving a tablet is

0.3*0.5 = 0.15 = 15%

Therefore, the probability of both children receiving a tablet is 15%

Therefore option b is correct.

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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]: