Answer is in the file below
tinyurl.com/wpazsebu
Number 1 is x=8!!
2 is pretty confusing so I hope someone else can answer two as I have answered 1!
Answer:
x = 107
Step-by-step explanation:
x and 107 are alternate interior angles and alternate interior angles are equal
so x= 107
Answer:
0.44
Step-by-step explanation:
Given the estimated logistic regression model on risk of having squamous cell carcinoma
-4.84 + 4.6*(SMOKER)
SMOKER = 0 (non-smoker) ; 1 (SMOKER)
What is the predicted probability of a smoker having squamous cell carcinoma?
exp(-4.84 + 4.6*(SMOKER)) / 1 + exp(-4.84 + 4.6*(SMOKER))
SMOKER = 1
exp(-4.84 + 4.6) / 1 + exp(-4.84 + 4.6)
exp^(-0.24) / (1 + exp^(-0.24))
0.7866278 / 1.7866278
= 0.4402863
= 0.44
Based on the information of the table, you have:
1. The ratio is 105/150. By simplifying you get for the ratio 7/10.
2. The students that prefer action movies are 75+90 = 165 and the total numbe of students is 180+240 = 420. Then, the fraction of students who prefer action movies is:
165/420 = 11/28
3. The fraction of seventh graders students that prefer action movies is:
75/180 = 5/12
4. The percent of student that prefer comedy is:
105 + 150 = 255 total student that prefer comedy
420 total number of students
the fraction is:
(x/100)420 = 255
solve for x:
x = 255(100/420)
x = 60.71
the percent of students is 60.71%
5. The percent of eighth graders student who prefer action moveis is:
(x/100)240 = 90
x = 90(100/240)
x = 37.5
the percent of students is 37.5%
6. To determine which from the given grades has the greatest percent of student that prefer action movies, calculate the percent of student in seventh-grade:
(x/100)180 = 75
x = 75(100/180)
x = 41.66
the percent of student is 41.66%
then, seventh grade has the greatest percent of student that prefer action movies.
