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Ivenika [448]
3 years ago
8

Least:

Mathematics
1 answer:
-Dominant- [34]3 years ago
5 0
Least- 2

Greatest- 12

Median- 7

Lower Q- 4

Upper Q- 8

Range-10

IQR- 4
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Question 1
KatRina [158]

Answer:

x2+x-4=0

Step-by-step explanation:

7 0
3 years ago
A rack of 15 billiard balls is shown. If one ball is selected at​ random, determine the odds against it containing a number grea
andre [41]

Using the probability and odds concepts, it is found that the odds against it containing a number greater than or equal to 7 is \frac{2}{3}.

  • A probability is the <u>number of desired outcomes divided by the number of total outcomes</u>.
  • An odd is the <u>number of desired outcomes divided by the number of non-desired outcomes</u>.

In the rack, there are 15 balls, numbered from 1 to 15. Of those, <u>6 are less than 7</u>(against it containing a number greater than or equal to 7 is equivalent to it containing a number less than 7), thus:

  • There are 6 desired outcomes.
  • There are 9 non-desired outcomes.

The odd is:

\frac{6}{9} = \frac{2}{3}

The odds against it containing a number greater than or equal to 7 is \frac{2}{3}.

A similar problem is given at brainly.com/question/21094006

7 0
3 years ago
A small regional carrier accepted 23 reservations for a particular flight with 20 seats. 14 reservations went to regular custome
MrRissso [65]

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

7 0
3 years ago
1. The lease common multiple of 3, 4, 6, and 8 is<br> OA. 8.<br> OB. 24.<br> O C.72.<br> OD.96.
almond37 [142]
The answer is ob.24
3 0
3 years ago
Read 2 more answers
Write the following:<br>a) 16 as a fraction of 152<br>b) 15 as a fraction of 210<br>​
Nesterboy [21]

Answer:

\frac{16}{152}

\frac{15}{210}

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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