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alex41 [277]
3 years ago
7

HElp wHat iS iT!?!?!??!

Mathematics
1 answer:
fomenos3 years ago
8 0

Answer:

a

Step-by-step explanation:

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What is the solution of the system?
mario62 [17]
Y + 7 = 3x.....y = 3x - 7

6x - 2y = 12
6x - 2(3x - 7) = 12
6x - 6x + 14 = 12
14 = 12 (incorrect)

when ur variables cancel out and u r left with an incorrect statement, this means there is no solution because ur lines are parallel

so ur answer is : no solution
5 0
3 years ago
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Somebody please help and explain it.
s2008m [1.1K]

Explanation is in the file

tinyurl.com/wpazsebu

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2 years ago
Jay gas $8 he wants to buy a book gir $7.50 the sales tax 6% does he have enough money
Reil [10]
6%=0.06
if you want to know how much he paid for the sales taxes you multiply 7.5 by 0.06 
7.5 . 0.06= 0.45
7.5+0.45=7.95
8>7.95
Jay have enough money to buy the book
8 0
2 years ago
What is 607/100 as a percent?
777dan777 [17]


607 / 100 = 6.07

6.07 = decimal form

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6 0
2 years ago
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The SAT Reasoning Test (formerly called the Scholastic Aptitude Test) is perhaps the most widely used standardized test for coll
slamgirl [31]

Answer:

The score is  x  =  1884

Step-by-step explanation:

From the question we are told that

     The population mean is  \mu  = 1500

     The standard deviation is  \sigma  =  300

     

From the question we are told that the score follow a normal distribution

i.e     X  \~  \   N( 1500 , 300)

The proportion of score in the top 10% is mathematically

           P(X > x )  =  P( \frac{X -  \mu}{\sigma }  > \frac{x -  \mu}{\sigma }   ) = 0.10

Where x is the minimum score required to be in the top 10%

Now the \frac{X -  \mu}{\sigma }   =  Z (The  \ Standardized \ value \  of  \  X)

  So

            P(X > x )  =  P( Z > \frac{x -  \mu}{\sigma }   ) = 0.10

So

            P(X > x )  =  P( Z > \frac{x -  1500}{300}   ) = 0.10

So the critical value of  0.10  from the normal distribution table is  Z_{0.10} =  1.28

So

               \frac{x -  1500}{300}   = 1.28

              x  =  1884

           

       

6 0
3 years ago
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