Answer:
I'm pretty sure he or she are right^^
Step-by-step explanation:
i dont know
EFSSSSSSSSSSSSSSSSSSSSSSSSS
1.67 cups.
10 divided by 6, then round.
Let us model this problem with a polynomial function.
Let x = day number (1,2,3,4, ...)
Let y = number of creatures colled on day x.
Because we have 5 data points, we shall use a 4th order polynomial of the form
y = a₁x⁴ + a₂x³ + a₃x² + a₄x + a₅
Substitute x=1,2, ..., 5 into y(x) to obtain the matrix equation
| 1 1 1 1 1 | | a₁ | | 42 |
| 2⁴ 2³ 2² 2¹ 2⁰ | | a₂ | | 26 |
| 3⁴ 3³ 3² 3¹ 3⁰ | | a₃ | = | 61 |
| 4⁴ 4³ 4² 4¹ 4⁰ | | a₄ | | 65 |
| 5⁴ 5³ 5² 5¹ 5⁰ | | a₅ | | 56 |
When this matrix equation is solved in the calculator, we obtain
a₁ = 4.1667
a₂ = -55.3333
a₃ = 253.3333
a₄ = -451.1667
a₅ = 291.0000
Test the solution.
y(1) = 42
y(2) = 26
y(3) = 61
y(4) = 65
y(5) = 56
The average for 5 days is (42+26+61+65+56)/5 = 50.
If Kathy collected 53 creatures instead of 56 on day 5, the average becomes
(42+26+61+65+53)/5 = 49.4.
Now predict values for days 5,7,8.
y(6) = 152
y(7) = 571
y(8) = 1631
Answer:
Step-by-step explanation:
Hello!
You have two populations of interest and want to compare them. If you define the study variables as:
X₁: average hourly wages of an employee of the Downtown store.
n₁= 25
X[bar]₁= $9
S₁= $2
X₂: average hourly wages of an employee of the North Mall store.
n₂= 20
X[bar]₂= $8
S₂= $1
Both samples taken are independent, assuming that both populations are normal and that their population variances are equal I'll use the Student's-t statistic with a pooled sample variance to calculate the Confidence interval:
95% CI for μ₁ - μ₂
(X[bar]₁-X[bar]₂) ± 
Sa= 1.64
(9-8)±2.017*
[0.007636;1.9923]
I hope it helps!
