Unit 2 Exam
Use the product rule to nd the derivave of le fullnwhg. cfw_Hint Wte 1he quantjr as a product]
kit = cfw_F 3?
km = 4m" a)
Unit 2 Exam
@12211
Fuel 1119 derivave of H19 function.
n II
_Tx1
Elf'2?
51
(ZN? '
The devave is 3 = Unit 2 Exam
n I
Unit2 Quiz1 (12.1 12.2)
! F'nd lhe derivalive onhe function.
*2
. 3r=x3+?x+5
"
I
y'=3)(2+?
Unit2 Quiz1 (12.1 12.2) Unit2 Quiz1 (12.1 12.2)
Explain the relationship between the slope and the derivative of x at x = a.
Choose the oorrect answer below.
[2
Review Ouiz:Unit1 Quiz1 (10.1 11.1 11.2)
Score:00tlpt -( 1aot15v E]
11.2.21
(3) Graph the given function, (b) nd all values of): where the function is discontinuous, and (6) nd the limit from the left and the right at any values ofx where the function is
Review Quiz:Unit1 Quiz1 (10.1 11.1 11.2)
Score: 0.5 0H pt 4 130f157 Test Score: 80%, 12of 1!
@ 11.2.21
(3) Graph the given function, (b) nd all values of x where the function is discontinuous, and (c) nd the limit from the left and the right at any value
Review Test: Unit 1 Exam
Score:of1pt I 1or157
10.1.23
Give the domain of the funeon dened below.
tcfw_x=[x2]ll2
The domain is [Zuni
[Simplify your answer. Type 1,reur answer in interval notaon.)
Review Test: Unit 1 Exam
Score: 0.29 of 1 pt 4 2 of 15 7
Unit 1 Quiz 2 (11.3 11.4)
11.3.7i
th the average rate of change for the function over the given interval.
vzex hetweenx= 4andx= 2
The average rate of change of 3 between I = 4 and x: 2 is lHISES.
cfw_Round to four decimal places as needed.)
Unit 1 Quiz 2
Unit 2 Quiz 2 (12.3)
Wlite le given funclinn as me campusiliun of two functions.
y = 1" 12 + 5:
Choose the DDI'TECI answer below.
1
cfw_IF A. If x = and 99: = 12 + 5x, 1en yr = QEXH-
4;
1
t' :3 E. lfx= _.J'; and x: 12+5x.19"=9cfw_3(]-
1
t' :3 C. |ffcf
Unit2 Quiz1 (12.1 12.2)
Use the product rule to nd the derivative.
1' = [sz + 3cfw_3x 4
y=18x21x+9
Unit2 Quiz1 (12.1 12.2)
@1227
Use the product rule to nd the derivative of the fullnwhg.
v=tx+?l(?+4]
1 1
21 _ 49
jtr= ?x2+?x 2+4 Unit2 Quiz1 (12.1 12.2)
Childhood Obesity
NRS427VN
Sunni Douty
Childhood Obesity
Obesity is defined by the CDC
Today we will discuss complications of childhood obesity and implications for adulthood.
A child who is above the 85th percentile for weight (weight more than 80 percen
Name: Sunni Douty
Date: 11-16-2016
Topic One: Mean, Variance, and Standard Deviatio
Please type your answer in the cell beside the question.
Please type you answer in the cell beside the question.
1. Identify the sampling technique being used. Every 20th
Name:
Date:
Topic 2: Population and Sampling
Distribution Excel Worksheet
Directions: Answer all questions and submit to instructor at the end of Topic 2
1 A. For a normal distribution that has a mean of 100,
and a standard deviation of 8, determine the Z
Workbook exercises 11 and 16
1) Age, length of labor, income, return to work and the number of hours working per week.
2) Sample size, median standard deviation was used to describe the length of labor in the
experimental group and population, median and
Name: Sunni Douty
Hypothesis Testing
Answer each question completely to receive full credit
1. There is a new drug that is used to treat leukemia. The following data
represents the remission time in weeks for a random sample of 21
patients using the drug.
1. The r value between variables 4 and 9 is r = -0.32.
2. The negative r value indicates that a negative relationship exists when one variable increases and
the other variable decreases. This r value of r = -0.32 has moderate strength since 0.3 < r < 0.5,
ANSWERS TO EXERCISE 23
Question 1
The r value for the relationship between Hamstring strength index 60o and the Shuttle run test is
-0.149. This r value shows a weak correlation between the two variables, as it is less than the 0.3
threshold for significa
Sunni Douty
Exercise 36 Questions
1) The F value suggests there is a significant difference between the results of the control and
treatment groups. The P value of 0.005 is < the alpha of 0.05. This suggests that the groups are
significantly different and
Results
Variance: 218.1777777777776
Standard Deviation: 14.77084214856342
Explanation
1. First, I added up all of the numbers:
59 + 62 + 69 + 75 + 76 + 80 + 83 + 84 + 97 + 107 = 792
2. I squared the total, and then divided the number of items in the data