Answer:
3.843
Step-by-step explanation:
(3x1)= 1
(8x1/10= 0.8
(4x1/100)=0.04
(3x1/1000)=0.003
Then, add together to get 3.843.
Step-by-step explanation:
x is -9
hope it helps!:)
Answer:
1)
M: 15
T: 14
W: 19
Th: 20
F: 9
Sa: 9
Su: 5
2) 18 degrees
Step-by-step explanation:
The Correct option is 36.60
Step-by-step explanation:
The key to solving this question lies
around measurement
conversion,specifically converting
grams into its equivalence in
kilograms.
1000 grams equal one kilogram
9 grams=9/1000 kg
9 grams=0.009 kg
70 grams=70/1000 kg 1000 grams equal one kilogram
9 grams=9/1000 kg
9 grams=0.009 kg
70 grams=70/1000 kg
70 grams=0.07 kg
Julie pays=$600*0.009=$5.4
Jacques pays=$600*0.07=$42
Jacques pays $36.60 more ($42-
$5.4) than Julie paid
Option is wrong because that was Jacques pays=$600*0.07=$42
Jacques pays $36.60 more ($42-
$5.4) than Julie paid
Option is wrong because that was
what Julie paid
Option D is wrong because that was
what Jacques paid
Option B is obviously wrong
Answer:
![P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68](https://tex.z-dn.net/?f=P%28B%27%20%5Ccup%20A%27%29%20%3D%20P%28%28A%20%5Ccap%20B%29%27%29%20%3D%201-P%28A%20%5Ccap%20B%29%3D%201-0.32%3D0.68)
See explanation below.
Step-by-step explanation:
For this case we define first some notation:
A= A new training program will increase customer satisfaction ratings
B= The training program can be kept within the original budget allocation
And for these two events we have defined the following probabilities
![P(A) = 0.8, P(B) = 0.2](https://tex.z-dn.net/?f=%20P%28A%29%20%3D%200.8%2C%20P%28B%29%20%3D%200.2)
We are assuming that the two events are independent so then we have the following propert:
![P(A \cap B ) = P(A) * P(B)](https://tex.z-dn.net/?f=%20P%28A%20%5Ccap%20B%20%29%20%3D%20P%28A%29%20%2A%20P%28B%29)
And we want to find the probability that the cost of the training program is not kept within budget or the training program will not increase the customer ratings so then if we use symbols we want to find:
![P(B' \cup A')](https://tex.z-dn.net/?f=%20P%28B%27%20%5Ccup%20A%27%29%20)
And using the De Morgan laws we know that:
![(A \cap B)' = A' \cup B'](https://tex.z-dn.net/?f=%20%28A%20%5Ccap%20B%29%27%20%3D%20A%27%20%5Ccup%20B%27)
So then we can write the probability like this:
![P(B' \cup A') = P((A \cap B)')](https://tex.z-dn.net/?f=%20P%28B%27%20%5Ccup%20A%27%29%20%3D%20P%28%28A%20%5Ccap%20B%29%27%29)
And using the complement rule we can do this:
![P(B' \cup A') = P((A \cap B)')= 1-P(A \cap B)](https://tex.z-dn.net/?f=%20P%28B%27%20%5Ccup%20A%27%29%20%3D%20P%28%28A%20%5Ccap%20B%29%27%29%3D%201-P%28A%20%5Ccap%20B%29)
Since A and B are independent we have:
![P(A \cap B )=P(A)*P(B) =(0.8*0.4) =0.32](https://tex.z-dn.net/?f=%20P%28A%20%5Ccap%20B%20%29%3DP%28A%29%2AP%28B%29%20%3D%280.8%2A0.4%29%20%3D0.32)
And then our final answer would be:
![P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68](https://tex.z-dn.net/?f=P%28B%27%20%5Ccup%20A%27%29%20%3D%20P%28%28A%20%5Ccap%20B%29%27%29%20%3D%201-P%28A%20%5Ccap%20B%29%3D%201-0.32%3D0.68)