Answer:
I
Step-by-step explanation:
graph is not provided
Answer:
$95.78
Step-by-step explanation:
f(t) = 300t / (2t² + 8)
t = 0 corresponds to the beginning of August. t = 1 corresponds to the end of August. t = 2 corresponds to the end of September. So on and so forth. So the second semester is from t = 5 to t = 10.
$T₂ = ∫₅¹⁰ 300t / (2t² + 8) dt
$T₂ = ∫₅¹⁰ 150t / (t² + 4) dt
$T₂ = 75 ∫₅¹⁰ 2t / (t² + 4) dt
$T₂ = 75 ln(t² + 4) |₅¹⁰
$T₂ = 75 ln(104) − 75 ln(29)
$T₂ ≈ 95.78
To solve this problem you must apply the proccedure shown below:
1- You must apply the following formula:
Where
is the residual, 
is the observed value 
and 
is the predicted value .
2- You only need to substitute the 
into the equation 
and then, you must apply the formula for calculate the residual:
The answer is: The residual is 
Answer:
Step-by-step explanation:
The given options are:
The key to determining which product is negative is to understand the rule of sign multiplication.
Now:
- The product of even negative terms is positive
- The product of odd negative terms is negative.
- The product of positive will always be positive.
In Options A and B, the number of negative signs is even, therefore our result is positive.
In option C, all the terms are positive, therefore our result will be positive.
In Option D, the number of negative signs is odd, therefore our result is negative.
Answer: 0.31
Step-by-step explanation:
Let A denotes the event of Tampa Bay Buccaneers will score a touchdown on their opening drive and B denote the event that their defense will have 3 or more sacks in the game.
Given : P(A)=0.14 P(B) = 0.31 P(A or B)=0.14
Formula : P(A and B)= P(A) + P(B) - P(A or B)
Now, the probability that they will both score a touchdown on the opening drive and have 3 or more sacks in the game will be :-
P(A and B)= 0.14 + 0.31 - 0.14=0.31
Hence, the required probability : 0.31
