Answer:
a) y = 4e^(0.06t)
b) 0.06
c) 11.55 hours
d) 6.9 million
Step-by-step explanation:
When the growth rate (millions per hour) is proportional to the number (millions), the relationship is exponential. The growth rate is the constant of proportionality.
a) Formula for y(t):
y = 4e^(0.06t)
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b) The growth constant is 0.06, the multiplier of t in the exponential function. It is the constant of proportionality in the given differential equation:
y' = 0.06y.
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c) The doubling time is found from ...
2 = e^(0.06t) . . . the multiplying factor is 2 to double the original number
ln(2) = 0.06t . . . . taking natural logs
ln(2)/0.06 = t ≈ 11.55 h . . . . doubling time
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d) Put t=9 into the formula from part (a). After 9 hours, there will be ...
y(9) = 4e^(0.06·9) ≈ 6.9 . . . . million bacteria present
Answer: it's C
Step-by-step explanation:
Given the function . The above function can be written as
a)Now, the function has minimum value since the coefficient of is .
b) The minimum value of the function occurs at and its value is
c)The minimum value of the function occurs at .
To answer this question, you need to understand how to change the description into an equation. Let say that Herbert's container volume is H and Betty's container is B. Then, from the problem you can get this equation:
1. B= 200 (Betty's container hold 200 milli liters)
2. H= 7/8 B (Herbert's container hold 7/8 as much juice)
If you put the 1st equation into the 2nd, you will get:
H=7/8 (200)= 175
Then the two containers will hold
H+B= 200 + 175= 375milli liters
Answer:
Step-by-step explanation:
Test the claim that the mean GPA of night students is significantly different than the mean GPA of day students at the 0.02 significance level.
GPA-Night GPA-Day 3.15 3.47 3.68 3.49 3.34 3.07 3.07 3.31 3.31 3.28 3.2 3.05 3.07 3.12 2.8 3.5 3.04 3.04 3.13 3.19 3.54 3.26 3.24 3.31 3.02 3.45 3.44 2.9 2.79 2.76 3.18 2.69 2.99 3.46 3.28 3.09 3.16 2.72 3.08 3.14 2.47 3.08 3.03 2.99 3.02 3.11 2.58 2.84 3.36 2.99 3.04 2.99
(1) The null and alternative hypothesis would be:
a.H0:μN≥μD
H1:μN>μD.
b.H0:μN=μD
H1:μN≠μD.
c.H0:pN≥pD
H1:pN d.H0:pN=pD
H1:pN≠pD.
e.H0:μN≤μD
H1:μN<μD
f.H0:pN≤pD
H1:pN>pD.
(2) The test is:_______.
a. right-tailed.
b. two-tailed.
c. left-tailed.
(3) The sample consisted of 70 night students, with a sample mean GPA of 2.12 and a standard deviation of 0.08, and 70 day students, with a sample mean GPA of 2.08 and a standard deviation of 0.02