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2 years ago

Graph the line with the equation y=2/5x + 3.

Mathematics

2 answers:

shusha [124]2 years ago

5 0

Answer:

3r=2

Step-by-step explanation:

Maru [420]2 years ago

5 0

The line would go through the points (-7.5,0) and (0,3)

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Please need help on this

jenyasd209 [6]

Answer: do you need help with 1 and 2?

Step-by-step explanation:

3 0

3 years ago

Let f(x)= x^2-4x-c. Find a nonzero value of c such that f(c)=c.

Dimas [21]

Answer:

The nonzero value of c will be:

$c = 6$

Step-by-step explanation:

Given the function

$f\left(x\right)=\:x^2-4x-c$

$f\left(c\right)=\:c^2-4c-c$

$f(c) = c$

$c=\:c^2-4c-c$

switching the sides

$c^2-4c-c=c$

subtract c from both sides

$c^2-4c-c-c=c-c$

$c^2-6c=0$

$c\left(c-6\right)=0$

Using the zero factor principle

$\:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$c=0\quad \mathrm{or}\quad \:c-6=0$

so, the solutions to the quadratic equations are:

$c=0,\:c=6$

Therefore, a nonzero value of c will be:

$c = 6$

5 0

2 years ago

What is 2.1 with 1repeating as a mixed number

swat32

Answer:

2 1/10. is the answer i think

4 0

3 years ago

Find the range and interquartile range for the data represented by the box plot.

Anarel [89]

<u>Given</u>:

Given that the data are represented by the box plot.

We need to determine the range and interquartile range.

<u>Range:</u>

The range of the data is the difference between the highest and the lowest value in the given set of data.

From the box plot, the highest value is 30 and the lowest value is 15.

Thus, the range of the data is given by

Range = Highest value - Lowest value

Range = 30 - 15 = 15

Thus, the range of the data is 15.

<u>Interquartile range:</u>

The interquartile range is the difference between the ends of the box in the box plot.

Thus, the interquartile range is given by

Interquartile range = 27 - 18 = 9

Thus, the interquartile range is 9.

3 0

2 years ago

Read 2 more answers

The heights of adult males in the United States are approximately normally distributed. The mean height is 70 inches (5 feet 10

steposvetlana [31]

Using the normal distribution, it is found that the probability is 0.16.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem, the mean and the standard deviation are given by, respectively, $\mu = 70, \sigma = 3$ .

The proportion of students between 45 and 67 inches is the p-value of Z when <u>X = 67 subtracted by the p-value of Z when X = 45</u>, hence:

X = 67:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 70}{3}$

Z = -1

Z = -1 has a p-value of 0.16.

X = 45:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 70}{3}$

Z = -8.3

Z = -8.3 has a p-value of 0.

0.16 - 0 = 0.16

The probability is 0.16.

More can be learned about the normal distribution at brainly.com/question/24663213

3 0

2 years ago