Answer:
12.9
Step-by-step explanation:
30400=
30400=
\,\,22000e^{0.025t}
22000e
0.025t
Plug in
\frac{30400}{22000}=
22000
30400
=
\,\,\frac{22000e^{0.025t}}{22000}
22000
22000e
0.025t
Divide by 22000
1.3818182=
1.3818182=
\,\,e^{0.025t}
e
0.025t
\ln\left(1.3818182\right)=
ln(1.3818182)=
\,\,\ln\left(e^{0.025t}\right)
ln(e
0.025t
)
Take the natural log of both sides
\ln\left(1.3818182\right)=
ln(1.3818182)=
\,\,0.025t
0.025t
ln cancels the e
\frac{\ln\left(1.3818182\right)}{0.025}=
0.025
ln(1.3818182)
=
\,\,\frac{0.025t}{0.025}
0.025
0.025t
Divide by 0.025
12.9360062=
12.9360062=t
t = 12.9
12.9
Answer
a) y | p(y)
25 | 0.8
100 | 0.15
300 | 0.05
E(y) = ∑ y . p(y)
E(y) = 25 × 0.8 + 100 × 0.15 + 300 × 0.05
E(y) = 50
average class size equal to E(y) = 50
b) y | p(y)
25 | 
100 | 
300 | 
E(y) = ∑ y . p(y)
E(y) = 25 × 0.4 + 100 × 0.3 + 300 × 0.3
E(y) = 130
average class size equal to E(y) = 130
c) Average Student in the class in a school = 50
Average student at the school has student = 130
2 (1/4)= 2.25*100= 225%
the answer to the question is a.
Answer:
Here is the graph I made, starting from (-3,1) to (-2,-1) it goes down 2 units and over 1 unit.
Answer:
b+6=19
Step-by-step explanation:
by the words you can write your equation if you are looking to solve your answer would be 13
