12 = r - (34 - 2)
subtract he 2 from 34
12 = r - (32)
add 32 to each side
r = 44
Check:
12 = 44 - (34 - 2)
12 = 44 - 32
12 = 12 :)
Answer:
0.0273
Step-by-step explanation:
np n
10 100
9 100
11 100
7 100
3 100
12 100
8 100
4 100
6 100
11 100
pbar=sumnp/sumn
pbar=10+9+11+7+3+12+8+4+6+11/10+10+10+10+10+10+10+10+10+10
pbar=81/1000
pbar=0.081
nbar=sumn/k=1000/10=100




Standard deviation for p-chart=0.0273
Answer:


Step-by-step explanation:
We are given that there is an exponential decay.
Also, the decrease is of constant rate 7.9% i.e. 0.079 each year.
Since, the initial amount of the species is atleast 26 million.
Thus, the inequality for the corresponding model will be,
,
where t is the time period for the decay and P is the population.
Moreover, is is given that the population cannot be less than 2 million.
So, we get,
.
Hence, the inequalities to determine the possible number of insects over time are given by,

.
Answer:
Option A
then you would have an angle, a side and another angle (ASA) next to each other
Opt.B would be a second sude (SAS)
Opt.C is already given
Opt.D seems to be like (A-S-S). no pun intended.
A:
(f+g)(x)=f(x)+g(x)
(f+g)(x)=4x-5+3x+9
(f+g)(x)=7x+4
B:
(f•g)(x)=f(x)•g(x)
(f•g)(x)=(4x-5)(3x+9)
(f•g)(x)=12x^2-15x+36x-45
(f•g)(x)=12x^2+21x-45
C:
(f○g)(x)=f(g(x))
(f○g)(x)=4(3x+9)-5
(f○g)(x)=12x+36-5
(f○g)(x)=12x+31