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3 years ago

Graph each linear equation y equals negative 3x - 2

Mathematics

2 answers:

den301095 [7]3 years ago

4 0

Answer: y=-(3x-2)= -3x+2

The intercept with y is 2

The intercept with x is 2/3

emmasim [6.3K]3 years ago

4 0

View the attached image for the graph.

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A line passes through point A (14,21). A second point on the line has an x-value that is 125% of the x-value of point A and a y-

seropon [69]

Answer:

The equation of the line in point-slope form is $y-21 = - \frac{3}{2}\cdot (x-14)$ .

Step-by-step explanation:

According to the statement, let $A(x,y) = (14,21)$ and $B(x,y) = (1.25\cdot x_{A},0.75\cdot y_{A})$ . The equation of the line in point-slope form is defined by the following formula:

$y-y_{A} = m\cdot (x-x_{A})$ (1)

Where:

$x_{A}$ , $y_{A}$ - Coordinates of the point A, dimensionless.

$m$ - Slope, dimensionless.

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

In addition, the slope of the line is defined by:

$m = \frac{y_{B}-y_{A}}{x_{B}-x_{A}}$ (2)

If we know that $x_{A} = 14$ and $y_{A} = 21$ , then the equation of the line in point-slope form is:

$x_{B} = 1.25\cdot (14)$

$x_{B} = 17.5$

$y_{B} = 0.75\cdot (21)$

$y_{B} = 15.75$

From (2):

$m = \frac{15.75-21}{17.5-14}$

$m = -\frac{3}{2}$

By (1):

$y-21 = - \frac{3}{2}\cdot (x-14)$

The equation of the line in point-slope form is $y-21 = - \frac{3}{2}\cdot (x-14)$ .

5 0

3 years ago

Help me and thx alot

Svet_ta [14]

Answer:

easyyyyyyy

Step-by-step explanation:

table J!

8 0

3 years ago

Does the frequency distribution appear to have a normal distribution using a strict interpretation of the relevant criteria?Tem

Xelga [282]

Answer:

B. No, this distribution does not appear to be normal

Step-by-step explanation:

Hello!

To observe what shape the data takes, it is best to make a graph. For me, the best type of graph is a histogram.

The first step to take is to calculate the classmark`for each of the given temperature intervals. Each class mark will be the midpoint of each bar.

As you can see in the graphic (2nd attachment) there are no values of frequency for the interval [40-44] and the rest of the data show asymmetry skewed to the left. Just because one of the intervals doesn't have an observed frequency is enough to say that these values do not meet the requirements to have a normal distribution.

The answer is B.

I hope it helps!