Answer:
Step-by-step explanation:
If an exponential function is in the form of y = a(b)ˣ,
a = Initial quantity
b = Growth factor
x = Duration
Condition for exponential growth → b > 1
Condition for exponential decay → 0 < b < 1
Now we ca apply this condition in the given functions,
1).
Here, (1 + 0.45) = 1.45 > 1
Therefore, It's an exponential growth.
2).
Here, (0.85) is between 0 and 1,
Therefore, it's an exponential decay.
3). y = (1 - 0.03)ˣ + 4
Here, (1 - 0.03) = 0.97
And 0 < 0.97 < 1
Therefore, It's an exponential decay.
4). y = 0.5(1.2)ˣ + 2
Here, 1.2 > 1
Therefore, it's an exponential growth.
<span><span>C=(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span><span>C=(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span><span> and radius </span><span>rr</span>.
<span><span>(x−<span>x0</span><span>)2</span>+(y−<span>y0</span><span>)2</span>+(z−<span>z0</span><span>)2</span>=<span>r2</span></span><span>(x−<span>x0</span><span>)2</span>+(y−<span>y0</span><span>)2</span>+(z−<span>z0</span><span>)2</span>=<span>r2</span></span></span><span> </span>
Answer:
P(A)=0.55
P(A and B)=P(A∩B)=0.1265
P(A or B)=P(A∪B)=0.7635
P(A|B)=0.3721
Step-by-step explanation:
P(A')=0.45
P(A)=1-0.45=0.55
P(B∩A)=?
P(B|A)=0.23
P(B|A)=(P(A∩B))/P(A)
0.23=(P(A∩B))/0.55
P(A∩B)=0.23×0.55=0.1265
P(A∪B)=P(A)+P(B)-P(A∩B)
=0.55+0.34-0.1265
=0.7635
P(A|B)=[P(A∩B)]/P(B)=0.1265/0.34 ≈0.3721
0 solutions
move the x to one side, and the numbers to the other.
12x (-12x) + 32 (-32)= 12 (-12x) - 7 (-32)
12x - 12x = -7 - 32
0 = -39
0 ≠ -39
∴ you have 0 solutions
hope this helps