The lest common multiple of 19 and 14 is 266
hope this helped
Answer:
They purchased 6 dozen eggs
Step-by-step explanation:
The total cost is the cost per dozen times the number of dozen purchased
total cost = 1.98 * n where n is the number of dozen
11.88 = 1.98n
Divide each side by 1.98
11.88/1.98 = 1.98/1.98 * n
6=n
They purchased 6 dozen eggs
<u>Answer:</u>
The value of x is in the solution set of 3(x – 4) ≥ 5x + 2 is -10
<u>Solution:</u>
Need to determine which value of x from given option is solution set of 3(x – 4) ≥ 5x + 2
Lets first solve 3(x – 4) ≥ 5x + 2
3(x – 4) ≥ 5x + 2
=> 3x – 12 ≥ 5x + 2
=> 3x – 5x ≥ 12 + 2
=> -2x ≥ 14
=> -x ≥ 7
=> x ≤ -7
All the values of x which are less than or equal to -7 is solution set of 3(x – 4) ≥ 5x + 2. From given option there is only one value that is -10 which is less than -7
Hence from given option -10 is solution set of 3(x – 4) ≥ 5x + 2.
4, 5, 3, 3, 1, 2, 3, 2, 4, 8, 2, 4, 4, 5, 2, 3, 6,2
Anna007 [38]
Answer:
<u>Given data:</u>
- 4, 5, 3, 3, 1, 2, 3, 2, 4, 8, 2, 4, 4, 5, 2, 3, 6,2
<u>Put the data in the ascending order:</u>
- 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 8
<u>Mean, the average:</u>
- (1 + 2*5 + 3*4 + 4*4 + 5*2 + 6 + 8)/18 = 3.5
<u>Median, average of middle two numbers:</u>
<u>Mode, the most repeated number:</u>
Mean is normally the best measure of central tendency, same applies to this data.
Answer:
- The probability that overbooking occurs means that all 8 non-regular customers arrived for the flight. Each of them has a 56% probability of arriving and they arrive independently so we get that
P(8 arrive) = (0.56)^8 = 0.00967
- Let's do part c before part b. For this, we want an exact booking, which means that exactly 7 of the 8 non-regular customers arrive for the flight. Suppose we align these 8 people in a row. Take the scenario that the 1st person didn't arrive and the remaining 7 did. That odds of that happening would be (1-.56)*(.56)^7.
Now take the scenario that the second person didn't arrive and the remaining 7 did. The odds would be
(0.56)(1-0.56)(0.56)^6 = (1-.56)*(.56)^7. You can run through every scenario that way and see that each time the odds are the same. There are a total of 8 different scenarios since we can choose 1 person (the non-arriver) from 8 people in eight different ways (combination).
So the overall probability of an exact booking would be [(1-.56)*(.56)^7] * 8 = 0.06079
- The probability that the flight has one or more empty seats is the same as the probability that the flight is NOT exactly booked NOR is it overbooked. Formally,
P(at least 1 empty seat) = 1 - P(-1 or 0 empty seats)
= 1 - P(overbooked) - P(exactly booked)
= 1 - 0.00967 - 0.06079
= 0.9295.
Note that, the chance of being both overbooked and exactly booked is zero, so we don't have to worry about that.
Hope that helps!
Have a great day :P
