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2 years ago

PLEASE ASAP! I will give brainliest!!

Mathematics

2 answers:

Orlov [11]2 years ago

7 0

The answer is YES

Step-by-step explanation:

PLS give brainliest

Kay [80]2 years ago

3 0

Answer:

Yes

Step-by-step explanation:

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Calculate the percent error. Jerri estimated that 30 people would attend the

kipiarov [429]

Answer:she errored

Step-by-step explanation:

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2 years ago

The measured weights of 1,000 men in a certain village follow a normal distribution with a mean of 150 pounds and a standard dev

earnstyle [38]

Z = (X-Mean)/SD
z1 = (165 - 150)/15 = +1 
z2 = (135 - 150)/15 = - 1 
According to the Empirical Rule 68-95-99.7 
Mean +/- 1SD covers 68% of the values 
100% - 68% = 32% 
The remaining 32% is equally distributed below z = - 1 and z = +1 
32%/2 = 16%

Therefore,

a) Number of men weighing more than 165 pounds = 16% of 1000 = 160 
b) Number of men weighing less than 135 pounds = 16% of 1000 = 160

4 0

2 years ago

Read 2 more answers

In the case of classical activation, which outcome is likely to occur?.

adell [148]

Classical activation should lead to an effective immune response against pathogens.

<h3>What is the classical activation pathway?</h3>

This is the most common pathway in the immune system.

<h3>What does classical activation imply?</h3>

This is triggered by specific antibody complexes that react to a potential pathogen.
A set of proteins and specialized cells are activated to attack the pathogen.
A membrane attack complex is created.

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

The desirable outcome of this system is an effective immune response against pathogens such as viruses and bacteria.

Learn more about pathogens in: brainly.com/question/13873423

6 0

1 year ago

I’ll give brainliest (worth 15 pts)

mixas84 [53]

Answer:

Step-by-step explanation:

( it might be wrong pls dont report me just let me kno y its wrong )

3 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]: