Y=10
Y=7x
10=7x
7/10= 1.43
A.$1.43
1.43•5=$7.15
B.$7.15
I think this is the answer hope it helps
Answer:
here i finished!
hope it helps yw!
Step-by-step explanation:
The doubling period of a bacterial population is 15 minutes.
At time t = 90 minutes, the bacterial population was 50000.
Round your answers to at least 1 decimal place.
:
We can use the formula:
A = Ao*2^(t/d); where:
A = amt after t time
Ao = initial amt (t=0)
t = time period in question
d = doubling time of substance
In our problem
d = 15 min
t = 90 min
A = 50000
What was the initial population at time t = 0
Ao * 2^(90/15) = 50000
Ao * 2^6 = 50000
We know 2^6 = 64
64(Ao) = 50000
Ao = 50000/64
Ao = 781.25 is the initial population
:
Find the size of the bacterial population after 4 hours
Change 4 hr to 240 min
A = 781.25 * 2^(240/15
A = 781.25 * 2^16
A= 781.25 * 65536
A = 51,199,218.75 after 4 hrs
Answer:
G oo g l ee I use it there's a scan
C would equal 20. because 12 times its self plus 16 times its self equals 400. then I square 400 and got 20, which equals c
Answer
There are 170 ways the student can answer the test.
Explanation
If there are 20 multiple-choice questions with 5 choices each, the student has 100 choices. The first question has 5 choices to pick from. The second has 5 as well. So does the third. Hopefully now you realize that you have to multiply the number of choices by the number of questions.
The same thing goes with the true/false questions. There are 2 choices for each true/false question, and there are 35 of those. 35×2 is 70. There are 70 ways to answer on the true/false questions.
Now combine the number of choices on the first part and the second part; 100+70 is 170.
