All categories

3 years ago

State the discriminant and the nature of the solution it will

Mathematics

1 answer:

____ [38]3 years ago

3 0

The discriminant is b² - 4ac, in this case equal to 16 + 84 = 100.

The nature of the solution is real and rational (if that's what you mean).

i hope this was helpful! :D

You might be interested in

Your family plans to start a small business in your neighborhood.Your father borrows $10,000 from the bank at an annual interest

Temka [501]

Answer:

10000*8%*3

=$2400 loan interest

4 0

3 years ago

Write 4.4354 correct to 2 decimal places

Andreyy89

4.4354 rounded to 2 decimal places is 4.44

6 0

3 years ago

Plz help, answer the question below

Afina-wow [57]

$\sqrt[6]{x^{16}} =\sqrt[6]{x^{12} \cdot x^{4}}=x^2\sqrt[6]{x^{4}}$

5 0

3 years ago

The mean annual income for people in a certain city is 37 thousand dollars, with a standard deviation of 28 thousand dollars. A

Aloiza [94]

Answer:

$P( 31 < \bar X< 41)$

And we can ue the z score formula given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got for the limits:

$z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515$

$z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01$

So we want to find this probability:

$P(-1.515$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the annual income of a population, and for this case we know the following info:

$\mu=37$ and $\sigma=28$ and we are omitting the zeros from the thousand to simplify calculations

We select a sample size of n=50>30.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by: