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dolphi86 [110]
3 years ago
9

How high would the rent have to be to qualify as an outlier

Mathematics
1 answer:
Akimi4 [234]3 years ago
8 0

Answer:

1500+

Step-by-step explanation:

You might be interested in
Are 15:35 and 25:45 equal
hammer [34]

Answer:No

Step-by-step explanation:

15:35

=15/35

=3/7 (cancellation)

while

25:45

=25/45

=5/8(cancellation)

therefore answer is no

6 0
3 years ago
Read 2 more answers
Which of the following expressions is equivalent to (13 + 2) (13 - 2)?
7nadin3 [17]

Answer:

d. 13^2 - 2^2

Step-by-step explanation:

The product of a sum and a difference is a difference of squares.

(13 + 2)(13 - 2) = 13^2 - 2^2

Answer: d. 13^2 - 2^2

8 0
3 years ago
Solve for x:<br> 6x + 6 = 3x – 21
Volgvan

Answer:

x = -9

Step-by-step explanation:

6x + 6 = 3x – 21

Subtract 3x from each side

6x-3x + 6 = 3x-3x – 21

3x+6 = -21

Subtract 6 from each side

3x+6-6 = -21-6

3x = -27

Divide by 3

3x/3 = -27/3

x = -9

6 0
4 years ago
2. Mark has scored 12, 14, 10, 8, and 5 points in the past 5 basketball games. Write and solve the inequality that
marusya05 [52]

Answer:

  Mark must score at least 11 points to have an average of 10 points per game.

Step-by-step explanation:

The average of a set of data is its sum, divided by the number of data points in the set. If p is the number of points Mark scores in his next game, and he wants his average at least 10, then the relation is ...

  (12 +14 +10 +8 +5 +p)/6 ≥ 10

  49 +p ≥ 60 . . . . . . . . . multiply by 6 and simplify

  p ≥ 11 . . . . . . . . . . . subtract 49

Mark must score at least 11 points in the next game to average at least 10 points per game.

_____

<em>Additional comment</em>

Sometimes, it is convenient to work with the average directly. Relative to the average Mark wants (10), his scores so far are +2, +4, 0, -2, -5, for a total of -1. In order to bring the average to the value Mark wants, he must bring this total to 0 or higher. That is, he must score at least 10+1 = 11 points in his next game.

5 0
3 years ago
There were 15 students running in a race. How many different arrangements of first, second, and third place are possible?
viktelen [127]

Answer:

The area 2730 possible arrangements of first, second, and third place.

Step-by-step explanation:

The order in which the students finish is important. For example A, B and C is a different finishing order(outcome) than B, A and C. So we use the permutations formula to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

P_{(n,x)} = \frac{n!}{(n-x)!}

In this question:

3 from a set of 15. So

P_{(15,3)} = \frac{15!}{(15-3)!} = 2730

The area 2730 possible arrangements of first, second, and third place.

7 0
4 years ago
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