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ioda
3 years ago
12

Y = 6x + 2 y = 6x - 2 how many solutions does this equation have

Mathematics
1 answer:
Alona [7]3 years ago
4 0

Answer:

Bro, just use photo math

Step-by-step explanation:

Photo math is free and will tell you the answer to any algebraic expression and explain how to work it out. this is my best answer. it's an app.

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Ira Lisetskai [31]

Step-by-step explanation:

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Entertainment Software Association would like to test if the average age of "gamers" (those that routinely play video games) is
lions [1.4K]

Answer:

"30 years or less when, in reality, the average age is more than 30 years"

Step-by-step explanation:

Type I error is produced when conclusion rejects a true null hypothesis.

The null hypothesis is

"The average gamer is more than 30 years old"

(deduced from the wording, not explicitly stated).

Then if the conclusion is "the average gamer is less than or equal to 30 years old" when in reality the average age is more than 30 years, then there is a type I error, since the null hypothesis is rejected.

Answer is D:

"30 years or less when, in reality, the average age is more than 30 years"

8 0
3 years ago
From 1900 to 2000 the average temperature in July in Atlanta was LaTeX: 90^\circ F90 ∘ F. A researcher believes due to Global Wa
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<em>*100% CORRECT ANSWERS</em>

7 0
3 years ago
How many different ways can you make 82 cents using current u.s. currency
andrew11 [14]
 <span>You can probably just work it out. 

You need non-negative integer solutions to p+5n+10d+25q = 82. 

If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80. 


So this is the same as n + 2d + 5q ≤ 16 

So now you simply have to "crank out" the cases. 

Case q=0 [ n + 2d ≤ 16 ] 

Case (q=0,d=0) → n = 0 through 16 [17 possibilities] 
Case (q=0,d=1) → n = 0 through 14 [15 possibilities] 
... 
Case (q=0,d=7) → n = 0 through 2 [3 possibilities] 
Case (q=0,d=8) → n = 0 [1 possibility] 

Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81 

Case q=1 [ n + 2d ≤ 11 ] 
Case (q=1,d=0) → n = 0 through 11 [12] 
Case (q=1,d=1) → n = 0 through 9 [10] 
... 
Case (q=1,d=5) → n = 0 through 1 [2] 

Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42 

Case q=2 [ n + 2 ≤ 6 ] 
Case (q=2,d=0) → n = 0 through 6 [7] 
Case (q=2,d=1) → n = 0 through 4 [5] 
Case (q=2,d=2) → n = 0 through 2 [3] 
Case (q=2,d=3) → n = 0 [1] 

Total from case q=2: 1 + 3 + 5 + 7 = 16 

Case q=3 [ n + 2d ≤ 1 ] 
Here d must be 0, so there is only the case: 
Case (q=3,d=0) → n = 0 through 1 [2] 

So the case q=3 only has 2. 

Grand total: 2 + 16 + 42 + 81 = 141 </span>
3 0
3 years ago
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