PLEASE ANSWER THIS QUESTION RIGHT PLEASE Devante Has a lunch account in the school cafeteria. His starting balance at the beginn
ing of the month is $35.50 The first week Devante But three lunches and his account balance was$28. The second week Devante  Bought five lunches and his account balance was $15.50 in a model that relates the balance of his lunch account as a function of the number of lunches he buys what does the rate of change represent? THE ANSWER CHOICES ARE IN THE PICTURE ABOVE
1 answer:
You might be interested in
We know that V = a^3 and that it's 1000 cubic meters (or unit cubes as long as one unit cube is 1x1x1m). We can write the calculation as attached below. The result (dimensions of one cube house) is 10x10x10 meters.
Answer:
Hence,we need at least 136 rainfall PH values in the sample i.e
n ≥ 136
Step-by-step explanation:
We are given that:
(σ1)^2 = (σ2)^2 = Population variance = 0.25
So, E < 0.1
Confidence coefficient (c) = 0.9
n = n1 = n2
For confidence level, 1 - α = 0.9,we'll determine Z (α /2) = Z 0.05 by looking up 0.005 using the normal probability table which i have attached.
So, Z (α /2) = 1.645
The margin of error E is given as;
E = Z (α /2)√[(σ1)^2)/n1] + [(σ2)^2)/n2]
= Z (α /2)√({(σ1)^2 + (σ2)^2}/n) < 0.1
Multiply both sides by √n to get;
Z (α /2)√(σ1)^2 + (σ2)^2} < 0.1√n
Divide both sides by 0.1;
{Z (α /2)√(σ1)^2 + (σ2)^2}}/0.1 <√n
When we square each side, we get
{Z (α /2)√(σ1)^2 + (σ2)^2}}/0.1} ^2 < n
We'll now fill in the known values and solve;
n > ( 1.645 x √{(0.25 + 0.25)/0.1}^2
n > 135.3 or approximately n > 136
Hence,we need at least 136 observations in the sample i.e
n ≥ 136
