Answer:
A = $1,650.24
Step-by-step explanation:
A = P(1 + r/n)^nt
Where
A = future value
P = present value = $1,600
r = interest rate = 3.1% = 0.031
n = number of periods = 4
t = time = 1 year
A = P(1 + r/n)^nt
= 1,600( 1 + 0.031 / 4)^4*1
= 1600( 1 + 0.00775)^4
= 1600(1.00775)^4
= 1600(1.0314)
= 1,650.24
A = $1,650.24
Answer:
at least 9 students in each cohort.
Step-by-step explanation:
Given that :
In a class, there are 25 students and each of them is either a sophomore, a freshman or a junior. We have to determine the number students in the same cohort.
Let us suppose there are equal number of students in each of the cohort.
Now let us assume that the number of the students in each cohort be 8, i.e. each as a freshman, a junior or a sophomore. Therefore, the total number in the all the cohorts will be 24 students only.
Thus, we can say that there are at least 
freshman, at least sophomore or at least junior in each of the cohort.
Answer:
The answer is 13! which equals 6227020800.
1) Solve the first inequality
3y ≤ 2x + 6
=> y ≤ 2x/3 + 2
2) Draw the line y = 2x/3 + 2
3) Shade the region below the line y = 2x/3 + 2
4) Solve the inequality 2x + y > 0
=> y > - 2x
5) draw the line y = -2x
6) Shade the region above the line y = -2x
7) The solution is the region between the two lines shaded including the portion of the line y = 2x/3 + 2
Answer:
Step-by-step explanation:
Hello!
What you want to test is if there is a difference between the number of courses taken by students of different years, you can determine 4 populations of students:
Population 1: Freshmen
Population 2: Sophomores
Population 3: Juniors
Population 4: Seniors
To each population you measured the study variable, setting 4 different variables. To compare these variables the best way is to make one test where you compare the four populations at the same time, so the hypothesis is:
H₀: μ₁ = μ₂ = μ₃ = μ₄
H₁: At least one of the population means is different from the others.
The most appropriate test to use to compare the population mean of the 4 groups is an ANOVA.
I hope you have a SUPER day!
