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Margarita [4]
3 years ago
9

Solve the following equation: 6X – 2 = -14

Mathematics
2 answers:
Tom [10]3 years ago
5 0
The Answer is: x= -2
lord [1]3 years ago
4 0
The answer would be -2
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In a recent 10-year period, the change in the number of visitors in the U.S. National parks was about -11,150,000. What was the
Trava [24]

Answer:

Mean = -1,115,000 per year.

Step-by-step explanation:

Mean = (Σx)/N

The mean is the sum of variables divided by the number of variables

Σx = sum of variable = change in number of visitors over a 10 year period = -11150000

N = number of variables = 10

Mean = (-11150000)/10

Mean = -1,115,000 per year.

Hope this Helps!!!

4 0
3 years ago
How would you solve (-1,4),y=3x+5
astraxan [27]

Answer:

y = 12x

Step-by-step explanation:

-1 times 4 equals -4 so you have -4,y = 3x + 5 so you do 5 + 3 so it is 8x then you have -4,y = 8x so you add 4 to each side and it becomes y = 12x

5 0
4 years ago
You are the treasurer for a local charity. At the beginning of the month the balance was $820.64. The charity received donations
djverab [1.8K]
Beginning Balance:                  820.64
Add:
donations: 500 + 55 + 25          580.00
Deduct:
electric bill:  40.64
postage bill: 12.75
rental       : 445.00            <u>     (498.39)</u>
Ending Balance                     902.25

The balance at the end of the month is $902.25
6 0
3 years ago
Help pls due soon!!!!!
S_A_V [24]
Each of the squares is 1 so that’s 37 plus 8 half’s which is 4 so 37 + 4 = 41
4 0
4 years ago
Read 2 more answers
I AM GIVING 45 POINTS TO WHOEVER GETS THIS RIGHT... plz answer corectly and try... i was working on this for sooo long. Plz try
Alexxx [7]
When you have 3 choices for each of 6 spins, the number of possible "words" is
  3^6 = 729

The number of permutations of 6 things that are 3 groups of 2 is
  6!/(2!×2!×2!) = 720/8 = 90

A) The probability of a word containing two of each of the letters is 90/729 = 10/81


The number of permutations of 6 things from two groups of different sizes is
  (2 and 4) : 6!/(2!×4!) = 15
  (3 and 3) : 6!/(3!×3!) = 20
  (4 and 2) : 15
  (5 and 1) : 6
  (6 and 0) : 1

B) The number of ways there can be at least 2 "a"s and no "b"s is
  15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.


_____
These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
6 0
3 years ago
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