Answer:
The probability that Jones is lying is 6/7
Step-by-step explanation:
First we will list out 2 different cases when the outcome is a lie
1.probability that Jones tells lies is = 0.6 and probability that dalgiliesh analyses it correctly is 0.8
So the probability that dagliesh correctly analyses that he is telling lies is 0.8*0.6=0.48
2.Probability that Jones tells truth is 0.4 and if dagliesh analyses it incorrectly (which has a probability of 0.2) the outcome(as analysed by dagiliesh) is a lie
So probability that dagliesh analyses Jones truth as a lie is 0.2*.0.4=0.08
Total probability of outcome being a lie is 0.48+0.08=0.56
But we need the probability of Jones actually saying a lie which is nothing but 0.48/0.56= 6/7
It would be 4x < h < 10x
the difference between 3x - 7x =4
and the sum would be 3x + 7x =10
Answer:
a) Number of laundry he can do with this box = 78 laundries.
b) Price he paying for each load = 0.256 $
Explanation:
Total scoop contained in detergent box = 195 scoops.
Scoops used by each load of laundry = 
scoops = 2.5 scoops
Number of laundry he can do with this box = 195/2.5 = 78 laundries.
Price of detergent box = 19.99 $
Price he paying for each load = Total price/ Number of loads
= 19.99/78
= 0.256 $
Answer:
This problem is incomplete, we do not know the fraction of the students that have a dog and also have a cat. Suppose we write the problem as:
"In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. X of the students who have a dog as a pet also have cat as a pet. If there are 45 students in her class, how many have both a dog and a cat as pets?"
Where X must be a positive number smaller than one, now we can solve it:
we know that in the class we have 45 students, and 4/5 of those students have dogs, so the number of students that have a dog as a pet is:
N = 45*(4/5) = 36
And we know that X of those 36 students also have a cat, so the number of students that have a dog and a cat is:
M = 36*X
now, we do not have, suppose that the value of X is 1/2 ("1/2 of the students who have a dog also have a cat")
M = 36*(1/2) = 18
So you can replace the value of X in the equation and find the number of students that have a dog and a cat as pets.
Answer:
A conjunction is true in region
✔ B <--- Answer
A disjunction is true in region(s)
✔ A, B, and C <--- Answer
A disjunction is false in region(s)
✔ D <--- Answer
Step-by-step explanation:
I took the assaignment aswell
