Answer:
All of the values in the data are used in calculating the mean.
The sum of the deviations is zero.
There is only one mean for a set of data.
Step-by-step explanation:
Required
True statement about arithmetic mean
(a) False
The mean can be equal to, greater than or less than the median
(b) True
The arithmetic mean is the summation of all data divided by the number of data; hence, all values are included.
(c) True
All mean literally represent the distance of each value from the average; so, when each value used in calculating the mean is subtracted from the calculated mean, then the end result is 0. i.e.
(d) True
The mean value of a distribution is always 1 value. When more values are added to the existing values or some values are removed from the existing values, the mean value will change.
(e) False
Nominal data are not numerical or quantitative data; hence, the mean cannot be calculated.
Answer:
2 (real) solutions.
Step-by-step explanation:
A quadratic always has two solutions, whether they are real or complex.
Sometimes the solution is complex, involving complex numbers (2 complex), sometimes they are real and distinct (2 real), and sometimes they are real and coincident (still two real, but they are the same).
In the case of
x^2+3x = 3, or
x² + 3x -3 = 0
we apply the quadratic formula to get
x = (-3 +/- sqrt(3^2+4(1)(3))/2
to give the two solutions
{(sqrt(21)-3)/2, -(sqrt(21)+3)/2,}
both of which are real.
There are 2 x intercepts, x=-4 and x=-1
<u> </u><u> </u><u> </u><u>–</u><u>3</u><u> </u><u> </u><u> </u><u>×</u><u> </u><u> </u><u> </u><u>–</u><u>4</u><u> </u><u> </u><u> </u><u> </u><u> </u>
10
<h3>HOPE THIS HELPS :)</h3>
C: none of these are solutions to the given equation.
• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.
• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.
The actual solution is easy to find, since this equation is separable.
<em>y'</em> - <em>y</em> = 0
d<em>y</em>/d<em>x</em> = <em>y</em>
d<em>y</em>/<em>y</em> = d<em>x</em>
∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>
ln|<em>y</em>| = <em>x</em> + <em>C</em>
<em>y</em> = exp(<em>x</em> + <em>C </em>)
<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>