Answer: 0.9332.
Step-by-step explanation:
Claim : College Algebra final exam score of engineering majors equal to 88.
Given that : The test statistic is z equals to 1.50.
To find the p-value (Probability value), we use standard normal distribution table, and search the p-value corresponds to the z-score.
In a Standard Normal Distribution Table below, the p-value corresponds z equals 1.5 is 0.9332.
Hence, the p-value is 0.9332.
Answer: Average rates of change for Bird A is 0.045 and for Bird B is 3.816.
Step-by-step explanation: Through its graph, a function can be analyzed and be identified its attributes. Average Rate of Change is the ratio of change in the function values to the change of x-value, i.e., it is the slope of the function in a specific interval of x. With the Average it is possible to compare two function.
<u>Average Rate of Change</u>
<em><u>Bird A</u></em>:
For the interval [0,18]:
x₁ = 0 f(0) = 8.3
x₂ = 18 f(18) = 9.1
Average =
Average =
Average = 0.045
<em><u>Bird B</u></em>: y = 3.6(1.06)x
For the interval [0,18]:
x₁ = 0 y = 3.6(1.06)0 = 0
x₂ = 18 y = 3.6(1.06)18 = 68.7
Average =
Average =
Average = 3.816
The Average Rate for Change for Bird A is 0.045 and for Bird B is 3.816. This means that the population of Bird B increase in rate faster than the population of Bird A.