Answer:
- Overall GPA=2.81
- It is not possible to get his GPA to 3.0 for graduation.
Step-by-step explanation:
The Student already has a GPA of 2.68 after 108 credit hours.
If he is taking 12 credit hours in his last semester and gets a perfect 4.0 GPA
Total Credits Earned Before = 2.68 X 108=2894.4
Projected Credit to be earned = 12 X 4= 48 Credits
Total credit Hour= 108+12=120 Hours
His Cumulative GPA = Total credits earned ÷ Total Credit Hour
Since 2.81 is less than 3.00, it is not possible to get his GPA to 3.0 for graduation.
Answer:
product A on machine 1 would make 27 units (rounded down) 81 dollars profit.
product A on machine 2 would make 25 units (rounded down) 75 dollars profit.
product b on machine 1 would make 15 units (rounded down) 75 dollars profit.
product b on machine 2 would make 16 units (rounded down) 80 dollars profit.
Step-by-step explanation:
a1 refers to product a on machine 1
a2 refers to product a on machine 2
same rule applies for product b 1 and 2
Answer:
D
Step-by-step explanation:
2x + 5y = - 10 Subtract 2x from both sides
5y = -2x - 10 Divide by 5
y = -2x/5 - 10/5
y = 0.4x - 2
The slope is minus so the line is going from lower right to upper left as B and D are doing.
The y intercept is - 2. B goes through + 2 so it is not the answer.
The answer is D
First we need to find k ( rate of growth)
The formula is
A=p e^kt
A future bacteria 4800
P current bacteria 4000
E constant
K rate of growth?
T time 5 hours
Plug in the formula
4800=4000 e^5k
Solve for k
4800/4000=e^5k
Take the log for both sides
Log (4800/4000)=5k×log (e)
5k=log (4800/4000)÷log (e)
K=(log(4,800÷4,000)÷log(e))÷5
k=0.03646
Now use the formula again to find how bacteria will be present after 15 Hours
A=p e^kt
A ?
P 4000
K 0.03646
E constant
T 15 hours
Plug in the formula
A=4,000×e^(0.03646×15)
A=6,911.55 round your answer to get 6912 bacteria will be present after 15 Hours
Hope it helps!
We look for the minimum of each function.
For f (x) = 3x2 + 12x + 16:
We derive the function:
f '(x) = 6x + 12
We match zero:
6x + 12 = 0
We clear the value of x:
x = -12/6
x = -2
We substitute the value of x in the equation:
f (-2) = 3 * (- 2) ^ 2 + 12 * (- 2) + 16
f (-2) = 4
For g (x) = 2sin(x-pi):
From the graph we observe that the minimum value of the function is:
y = -2
Answer:
A function that has the smallest minimum y-value is:
y = -2
