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aksik [14]
3 years ago
10

Which of the following sentences is correct?a. The greater the number of independent variables measured, the easier it is to int

erpret higher-order interactions.b. The greater the number of independent variables measured, the more difficult it is to interpret higher-order interactions.c. The greater the number of independent variables measured, the more difficult it is to interpret the lowest-order interaction(s).d. The greater the number of independent variables measured, the easier it is to interpret the lowest-order interaction(s)
Mathematics
1 answer:
Marat540 [252]3 years ago
8 0

Answer:

b. The greater the number of independent variables measured, the more difficult it is to interpret higher-order interactions.

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BRAINLIEST 20PTSSSS
Licemer1 [7]

Answer:

A,D,E

Step-by-step explanation:

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3 years ago
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PLEASE HELP!!!
Svetradugi [14.3K]
B or c for the equation because it is 1-n
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Use the substitution method to solve the system of equations. Choose the correct ordered pair. x + y = 11 -x = -y - 9
suter [353]
x+y=11 \\
-x=-y-9 \\ \\
\hbox{substitute the second equation by -1} \\ \\
x+y=11 \\
x=y+9 \\ \\
\hbox{substitute y+9 for x in the first equation} \\ \\
y+9+y=11 \\
2y=11-9 \\
2y=2 \\
y=\frac{2}{2} \\
y=1 \\ \\
x=y+9 \\
x=1+9 \\
x=10 \\ \\
(x,y)=(10,1)

The answer is B.
5 0
3 years ago
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Hurry show your work
Nikolay [14]

Answer: \huge\boxed{\boxed{15}}

Step-by-step explanation:

<u>Given expression</u>

\large\boxed{\frac{12[30 - (9+4^2)]}{|10|-|-6| } }

<u>Simplify the exponents</u>

\large\boxed{=\frac{12[30 - (9+16)]}{|10|-|-6| } }

Simplify values in the parenthesis

\large\boxed{=\frac{12[30 - 25]}{|10|-|-6| } }

\large\boxed{=\frac{12[5]}{|10|-|-6| } }

<u>Simplify absolute values (all positive)</u>

\large\boxed{=\frac{12[5]}{10-6 } }

\large\boxed{=\frac{12[5]}{4 } }

<u>Simplify by division</u>

\large\boxed{=3~[5]}

<u>Simplify by multiplication</u>

\huge\boxed{\boxed{=15}}

Hope this helps!! :)

Please let me know if you have any questions

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The price of the house is $500,000.
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