Answer:
C. (see the attachment)
Step-by-step explanation:
Both inequalities include the "or equal to" case, so both boundary lines will be solid. That excludes choices A and D.
The first inequality is plotted the same way in all graphs, so we must look at the second inequality. The relationship of y and the comparison symbol is ...
-y ≥ (something)
If we multiply by -1, we get ...
y ≤ (something else)
This means the solution space will be <em>on or below (less than or equal to) the boundary line</em>. This is the shaded area in graph C. (Graph B shows shading <em>above</em> the line.)
___
<em>Further comment</em>
Since the boundary for the second inequality is fairly steep, "above" and "below" the line can be difficult to see. Rather, you can consider the relationship of x to the comparison symbol. For the second inequality, that is ...
x ≥ (something)
indicating the solution space is <em>on or to the right of the boundary line</em>.
Answer:
0.96 in^3
Step-by-step explanation:
Are those the only three choices? Anyway, I hope this helps
We use the formula:
x± s <span>t_<span>(1−c)/2 / </span></span>√n = 5
(1 - c)/2 = (1 - 0.95) / 2 = 0.025
Using the t score calculator at this value, at the 95% confidence internval,
t = 0.5080, So
100 - 19(0.5080) / √n = 5
Solving for n
n = 97
a) 97
b) yes
Answer:
2nd option bro
Step-by-step explanation:
Answer:
4
Step-by-step explanation: 5 22 8 18 e
Graph the function and look for sign changes in the slope (easiest)
of the 7 roots 3 are real and 4 are complex
or the perferred way is to apply Descartes rule of signs
x^7-3x^5+4x^2-1=0 a typo in your function attached photos shows 4x²
(x^7) - (3x^5) +(4x^2) - (1) sign changes I see 3
then to find a bound on the number of negative roots
(-x^7) - (-3x^5) +(4x^2) - (1) = -x^7 + 3x^5 + 4x^2 - 1 I see 2 sign changes ??
there is at least 1 real root (due the odd highest power)
and either 0, 2, 4 or 6 complex roots
so 5 and 7 are out
still confused I used a polynomials root calculator
it said ther are three real roots (confirmed by the graph of the function)
and four comple roots (which I can't figure out from the graph)