Answer:
Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.
How long will it take for this population to grow to a hundred rodents? To a thousand rodents?
Step-by-step explanation:
Use the initial condition when dp/dt = 1, p = 10 to get k;
Seperate the differential equation and solve for the constant C.
You have 100 rodents when:
You have 1000 rodents when:
Answer: I feel the answer is C
Step-by-step explanation:
Y= the number riders over time
x= hours into Tara's shift
y= 33x + 132
<span>We are given f(1) = 0 and f(2) = 1.
Going forward, the term is the sum of the two previous terms.
f(3) = f(2) + f(1) = 0 + 1 = 1.
f(4) = f(3) + f(2) = 1 + 1 = 2
f(5) = f(4) + f(3) = 2 + 1 = 3
This matches answer B.</span>
Answer:
Step-by-step explanation:
The mean SAT score is 
, we are going to call it \mu since it's the "true" mean
The standard deviation (we are going to call it 
) is
Next they draw a random sample of n=70 students, and they got a mean score (denoted by 
) of 
The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.
- So the Null Hypothesis 
- The alternative would be then the opposite 
The test statistic for this type of test takes the form
and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.
With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.
<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>
