You're given the slope of the line and a point thru which the line passes. So it makes sense to use the point-slope form of the equation of a straight line: if the point is (a,b) and the slope is m, then

y-b=m(x-a). You'd then solve this for y to obtain the equation in slope-intercept form.

On the other hand, if you're asked to write the equation in slope-intercept form, starting with the general form of this equation may be faster: y=mx+b.

Substitute the given values for x and y (which are 25 and -9) and m (which is (2/5). Solve the resulting equation for b (the y-intercept).

Then write the finished equation: y=( ? )x + b, where b is the y-intercept you've just found.

8 0

3 years ago

The following table presents the time taken to review articles that were submitted for publication to a particular journal durin

belka [17]

A) Variance = 10.24

B) Standard Deviation = 3.2

One of the measurements of dispersion is the standard deviation, which is exclusively used for quantitative data. It aids in determining if the data's mean is a suitable measurement to reflect the core value.

TIME FREQUENCY(f) MIDPOINT(x) d d² fd fd²

0 - 0.9 43 0.45 -3 9 -129 387

1.0 - 1.9 17 1.45 -2 4 -34 68

2.0 - 2.9 19 2.45 -1 1 -19 19

3.0 - 3.9 18 3.45 0 0 0 0

4.0 - 4.9 14 4.45 1 1 14 14

5.0 -5.9 16 5.45 2 4 32 64

∑f = 127

∑fd = -136

∑fd² = 552

$d = \frac{x - 3.45}{1}$

Standard Deviation = $\sqrt \frac{fd^{2} }{N} - \sqrt \frac{(fd)^{2} }{N} * i$

$\sqrt \frac{552}{127} -\sqrt (\frac{-136}{127})^{2} * 1$

√4.35 - √1.15

Standard Deviation = 3.2

(SD)² = (3.2)² = 10.24

Variance = 10.24

To know more about standard deviation, refer to this link :

brainly.com/question/12402189

#SPJ1

7 0

1 year ago

identify the type of conic section that has the equation 4x^2+25y^2=100 and identify its domain and range.

Vika [28.1K]

4x^2+25y^2=100

$4x^2+25y^2=100 \\\\
Dividing \;both \;sides \;by \;100 \\\\
\frac{4x^2}{100}+ \frac{25y^2}{100}= \frac{100}{100} \\\\
\frac{x^2}{25}+ \frac{y^2}{4}= 1 \\\\
\frac{x^2}{5^2}+ \frac{y^2}{2^2}= 1 \\\\
This \;is \;an \;ellipse. \\\\ \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$

Domain of an ellipse is x ∈ [-a,a] and Range of an ellipse is y ∈ [-b,b].

So, given equation is an Ellipse where Domain is x ∈ [-5,5] and Range is y ∈ [-2,2].

4 0

2 years ago