The most general antiderivative of the given function g(t) is (8t + t³/3 + t²/2 + c).
The antiderivative of a function is the inverse function of a derivative.
This inverse function of the derivative is called integration.
Here the given function is: g(t) = 8 + t² + t
Therefore, the antiderivative of the given function is
∫g(t) dt
= ∫(8 + t² + t) dt
= ∫8 dt + ∫t² dt + ∫t dt
= [8t⁽⁰⁺¹⁾/(0+1) + t⁽²⁺¹⁾/(2+1) + t⁽¹⁺¹⁾/(1+1) + c]
= (8t + t³/3 + t²/2 + c)
Here 'c' is the constant.
Again, differentiating the result, we get:
d/dt(8t + t³/3 + t²/2 + c)
= [8 ˣ 1 ˣ t⁽¹⁻¹⁾ + 3 ˣ t⁽³⁻¹⁾/3 + 2 ˣ t⁽²⁻¹⁾/2 + 0]
= 8 + t² + t
= g(t)
The antiderivative of the given function g(t)is (8t + t³/3 + t²/2 + c).
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Answer:
C. Events E and A are independent
Step-by-step explanation:
we will verify each options
(a)
We can use independent events formula
P(B∩C)=P(B)*P(C)
we are given
P(B)=0.4
P(C)=0.25
P(B∩C)=0.05
now, we can plug these values into formula
and we get
0.05=0.4*0.25
0.05=0.1
we can see that left side is not equal to right side
so, this is FALSE
(b)
We can use independent events formula
P(D∩A)=P(D)*P(A)
we are given
P(D)=0.25
P(A)=0.6
P(D∩A)=0.1
now, we can plug these values into formula
and we get
0.1=0.25*0.6
0.1=0.15
we can see that left side is not equal to right side
so, this is FALSE
(c)
We can use independent events formula
P(E∩A)=P(E)*P(A)
we are given
P(E)=0.5
P(A)=0.6
P(E∩A)=0.3
now, we can plug these values into formula
and we get
0.3=0.5*0.6
0.3=0.3
we can see that both sides are equal
so, this is TRUE
(d)
We can use independent events formula
P(D∩B)=P(D)*P(B)
we are given
P(D)=0.25
P(B)=0.4
P(D∩A)=0.15
now, we can plug these values into formula
and we get
0.15=0.25*0.4
0.15=0.1
we can see that left side is not equal to right side
so, this is FALSE
1. 25 x 0.39 = $9.75
2. 5.56 + 5.81 = 11.37, 11.37/2 = 5.685
Answer:
idk but my guess is d
Step-by-step explanation:
Method 1 :
(152+138+160)÷3 = 150
(152+138+160+198)÷4 = 162
Then his average score will increase
Method 2 :
Since 198 > 150 then the average will increase automatically.