Kenji wants to divide 48.6 by 0.15. What is the first step that Kenji must complete?

If 5 (3x - 7) = 20 what is 6x - 8

garri49 [273]

You would have to solve for x the plug in the answer for c into the next equation to find your answer

6 0

3 years ago

What are the zeros of the function

aleksklad [387]

Answer:

Image result for What are the zeros of the function

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Description

DescriptionIn mathematics, a zero of a real-, complex-, or generally vector-valued function, is a member of the domain of such that vanishes at; that is, the function attains the value of 0 at, or equivalently, is the solution to the equation. A "zero" of a function is thus an input value that produces an output of

4 0

3 years ago

The following back to back stemplot represents the amount a sample of students spent on their last haircut. Because there are si

krek1111 [17]

The 5 - number summary of the distribution obtained from the boxplot is :

Minimum = 20
Maximum = 85
Lower quartile = 35
Median = 44
Upper quartile = 58
Outlier = 100

Firstly, we need to extract the values for the amount spent by females :

20, 25, 25, 35, 37, 38, 39, 43, 45, 46, 50, 53, 60, 70, 85, 100

The five number summary include ;

Minimum, lower Quartile, Median, Upper Quartile and Maximum values.

Rearranging the data in ascending order :

20, 25, 25, 35, 37, 38, 39, 43, 45, 46, 50, 53, 60, 70, 85, 100

The boxplot clearly and easily gives the five number summary of the distribution :

Minimum = 20 (starting point of the whisker)
Lower quartile = 35 (starting point of the box)
Median = 44 (vertical line in between the box)
Upper quartile = 58 (endpoint of the box)
Maximum = 85 (endpoint of the upper whisker).

Outlier :

Lower boundary = LOWER QUARTILE - (1.5 × IQR)
Upper boundary = UPPER QUARTILE + (1.5 × IQR)
IQR = UPPER QUARTILE - LOWER QUARTILE
IQR = (58 - 35) = 23

Lower boundary = 35 - (1.5 × 23) = 0.5

Upper boundary = 58 + (1.5 × 23) = 92.5

The point denoted by a dot on the boxplot is an outlier = 100

Hence, values lesser than 0.5 or greater than 92.5 are outliers.

Therefore, 100 is an outlier.

Learn more : brainly.com/question/24582786

7 0

3 years ago

What would this be?:)

Mnenie [13.5K]

Answer:

\d

Step-by-step explanation:

:)

3 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:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want to find this probability:

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

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:

$P(-1.515$

4 0

3 years ago