Your friend claims that any two arcs with the same measure are similar. Your cousin claims that any two arcs with the same measu
2 answers:
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This question is not complete. Find attached the complete question and the figure showing the two way table.
Answer:
a. P(owns a cellphone) = 781/1000
b. P(does not own a PS4) = 421/500
c. P(owns a PS4 U does not own a cellphone) = 377/1000
d. P(owns a PS4 n does not own a cellphone) = 25/158
e. P(owns a PS4 | own a cell phone) =
158/1000
f. P(owns a cellphone | own a PS4) = 781/1000
Step-by-step explanation:
Total number of students = 1000 students
We have two variables,
1. The students who own cellphones
2. The students who own PS4
a. P(owns a cellphone)
From the question and the table, we are told that 781 people own cellphones.
Hence, P(owns a cellphone) = 781/1000
b. P(does not own a PS4)
From the attached table, we can see that the number of students that does not own a PS4 = 842
Hen P(does not own a PS4) = 842/1000 = 421/500
c. P(owns a PS4 U does not own a cellphone) =
First we find the Probability of student that owns a PS4
Number of students that own a PS4 = 158
P(owns a PS4 ) = 158/1000
Number of students that does not own a cell phone = 219
P(does not own a cell phone) = 219/1000
Therefore, P(owns a PS4 U does not own a cellphone) = P(owns a PS4 ) + P(does not own a cell phone)
= 158/1000 +219/1000
= 377/1000
d. P(owns a PS4 n does not own a cellphone) =
From the table, we obtain that as =
25/158
e. P(owns a PS4 | own a cell phone) =
P(owns a PS4 n own a cell phone) ÷ P(own a cell phone)
= P(owns a PS4)
From the table that would give us = 158/1000
f. P(owns a cellphone | own a PS4) =
P(owns a cellphone n own a PS4) ÷ P(own a PS4)
= P(owns a cellphone
From the table, this is given as :
781/1000
Answer: y = -7
graph attached.
Step-by-step explanation:
This is a horizontal line passing through -7 in the y-axis.<u> It passes through </u><u>all x-values</u>
