Assuming that the inequality you were going for was a ≤, set both polynomials less than or equal to 0.
x - 3 ≤ 0
x + 5 ≤ 0
For the first equation add 3 to both sides of the inequality. For the second, subtract 5 from both sides.
x ≤ 3
x ≤ - 5
These would be your solutions I guess, however, if you want to expand upon that, your actual answer is (- ∞, - 5] because if you were to plot these two inequalities on a number line, that is where the overlap would occur.
Answer:
D.No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population.
Step-by-step explanation:
(a) Range is the difference between the smallest and largest observation.
Here Smallest observation = 0.63
and Largest observation = 1.48
⇒ Range = 0.85
(b) Standard Deviation is calculate by,

where,
is mean of the observation.
Here, Mean = 0.988
⇒ Standard Deviation = 0.313
(c) Variance is the square of Standard deviation.
Thus, Variance = (Standard Deviation)² = 0.098
(d) Here last option(D) is true i.e. No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population.
Your sequence
-8, 16/3, -32/9, 64/27
is a geometric sequence with first term -8 and common ratio
(16/3)/(-8) = (-32/9)/(16/3) = -2/3
The general term an of a geometric sequence with first term a1 and ratio r is given by
an = a1·r^(n-1)
For your sequence, this is
an = -8·(-2/3)^(n-1)
Step-by-step explanation:
Isolate the variable
10x - 7x = -20 - 8
3x = -28
x = -9.333333333333
Graph of f(x-3) is compressed by a factor of
horizontally of f(x).
<u>Step-by-step explanation:</u>
We have, the graph of f(x)=
, on replacing f(x) by f(x-3) we get:
=
.Below shown are the images for graph of f(x) and f(x-3). Both are functions are exponential , and so having exponential graph but f(x-3) is compressed by a factor of
horizontally . Domain and range of both functions are same i.e. F(x) & f(x-3) domain & range are same , just difference in graph :
.