Answer:
Step-by-step explanation:
We are given the following variables:
μ = the sample mean = 152 pounds
σ = the standard deviation = 26 pounds
x = the sample value we want to test = 180 pounds
n = the sample size = unknown
MOE = margin of error = 4% = 0.04
Confidence level = 96%
The first thing we can do is to find for the value of z
using the formula:
z = (x – μ) / σ
z = (180 – 152) / 26
z = 1.0769 = 1.08
Since we are looking for the people who weigh more than
180 pounds, therefore this is a right tailed z test. The p value is:
p = 0.1401
Then we can use the formula below to solve for n:
n = z^2 * p * (1 – p) / (MOE)^2
n = 1.08^2 * 0.1401 * (1 – 0.1401) / (0.04)^2
n = 87.82 = 88
Therefore around 88 people must be surveyed.
Answer:
1. The solutions marked on the attached graph
2.The solutions are (-2, 0) and (6, 0)
3. The discriminant would be positive
Step-by-step explanation:
- The solution of the quadratic equation is the values of x at y = 0, which means the points of intersection of the graph and the x-axis
- The discriminant is positive if the graph intersects the x-axis at two different points
- The discriminant is zero if the graph intersects the x-axis at only one point
- The discriminant is negative if the graph does not intersect the x-axis
<em>Based on the given graph</em>
1.
Look at the attached figure to see the solutions
2.
∵ The graph intersects the x-axis at points (-2, 0), (6, 0)
∴ The solutions are (-2, 0) and (6, 0)
3.
∵ The discriminant is positive if the graph intersects the x-axis
at 2 different points
∵ The given graph intersects the x-axis at points (-2, 0) and (6, 0)
→ They are different points
∴ The discriminant would be positive
Answer:
Increases
Step-by-step explanation:
Increasing interval goes up and up gradually. So now as x goes to the right, y should go up too. This increases...
So, let planet X's orbital period be T and its mean distance from the sun be A. Also let planet Y's orbital period be T_1, so that means if planet Y's mean distance from the sun were double that of planet X:
Which means that the orbital period in planet Y is increased by a factor of