Answer:
3.85 hours
Step-by-step explanation:
We have that the model equation in this case would be of the following type, being "and" the concentration of bacteria:
y = a * e ^ (b * t)
where a and b are constants and t is time.
We know that when the time is 0, we know that there are 100,000 bacteria, therefore:
100000 = a * e ^ (b * 0)
100000 = a * 1
a = 100000
they tell us that when the time is 2 hours, the amount doubles, that is:
200000 = a * e ^ (b * 2)
already knowing that a equals 100,000
e ^ (b * 2) = 2
b * 2 = ln 2
b = (ln 2) / 2
b = 0.3465
Having the value of the constants, we will calculate the value of the time when there are 380000, that is:
380000 = 100000 * (e ^ 0.3465 * t)
3.8 = e ^ 0.3465 * t
ln 3.8 = 0.3465 * t
t = 1.335 / 0.3465
t = 3.85
That is to say that in order to reach this concentration 3.85 hours must pass
Answer:
Step-by-step explanation:
245000 last year
This year 25235
(y2 - y1) / y1)*100 = your percentage change
(where y1=start value and y2=end value)
(( £25.235 - £24.500) / £24.500) * 100 = 0 %
There ain't no percentage change as there needs to be a bigger difference between the two numbers plus u should use the formula
Answer:
The hours when zero calls were received were most possibly when Madera turned the phone off while playing in a soccer game.
Step-by-step explanation:
According to the question,
Madera, a middle school student, received some phone calls on one Saturday from 10 a.m. to 10 p.m.
The hours when zero calls were received were most possibly when Madera turned the phone off while playing in a soccer game.
Answer:
their sum: 2.8m
their difference: 1.2m
the third side's length should be smaller than their sum, larger than their difference
it could be: 1.3-1.4-1.5-1.6-1.7-1.8-1.9-2-2.1-2.2-2.3-2.4-2.5-2.6-2.7 but since it has to be a whole number, 2m is the only eligible answer.
You haven't provided the choices, therefore, I cannot provide an exact answer. However, I will help you with the concept.
For an order pair to be a solution to a system of equations, it has to satisfy <u>BOTH</u> equations. If it satisfies only one equation of the system or satisfy neither of the equations, the, it is not a solutions
<u><em>Examples:</em></u>
<u>System 1:</u>
x = y + 1
2x + 3y = 7
Let's check (2,1)
2 = 1 + 1 ........> equation 1 is satisfied
2(2) + 3(1) = 7 ......> equation 2 is satisfied
<u>(2,1) is a solution to this system</u>
<u>System 2:</u>
y = x + 3
y = x - 1
Let's check (2,1):
1 ≠ 2 + 3 ........> equation 1 isn't satisfied
1 = 2 - 1 ..........> equation 2 is satisfied
<u>(2,1) isn't a solution to this system</u>
<u>System 3:</u>
2y = 9 - 3x
3x + 2y = 9
Let's ceck (2,1):
2(1) ≠ 9 - 3(2) ..........> equation 1 isn't satisfied
3(2) + 2(1) ≠ 9 .........> equation 2 isn't satisfied
<u>(2,1) isn't a solution to this system
</u>
<u><em>Based on the above,</em></u> all you have to do is substitute with (2,1) in the system you have and pick the one where both equations are satisfied
Hope this helps :)
