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romanna [79]
3 years ago
10

Mark just received word that he scored 88% on a trigonometry test. Mark is also in an advanced algebra course. Is Mark's score o

n his trigonometry test an accurate predictor of how well students in the trigonometry class will perform on the same test? A.No, because Mark is not representative of the population of the trigonometry class B.Yes, because Mark is representative of the population of the trigonometry class C.No, because Mark scored too high on the exam D.Yes, because Mark scored very high on the test, so it must have been an easy exam
Mathematics
1 answer:
katen-ka-za [31]3 years ago
4 0
First off, it would not be C because there are no guidelines to score too high on a test
Then, D would also be wrong because since he scored high, it doesn't just mean that the test was easy.
Now it comes to the yes or no between A and B, since he is just one of the many students in Trigonometry, it cant be a very accurate representation therefore the answer to the problem would be A
Solution: A. No, because Mark is not representative of the population of the trigonometry class
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Describe and correct the error in solving the inequality 9x > -45
Mnenie [13.5K]
9x > -45
__ __
9 9

x > -5-----> ANSWER


if :
-9x > -45
___ ___
-9 -9

x < 5
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Need help with my homework ​
Volgvan

Answer:

\displaystyle y=\frac{16-9x^3}{2x^3 - 3}

\displaystyle y=-\frac{56}{13}

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3

Simplify the radical with the cube:

\displaystyle x^3=\frac{3y+16}{2y+9}

Multiply by 2y+9

\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)

Simplify:

\displaystyle x^3(2y+9)=3y+16

Operate the parentheses:

\displaystyle x^3(2y)+x^3(9)=3y+16

\displaystyle 2x^3y+9x^3=3y+16

Subtract 3y and 9x^3:

\displaystyle 2x^3y - 3y=16-9x^3

Factor y out of the left side:

\displaystyle y(2x^3 - 3)=16-9x^3

Divide by 2x^3 - 3:

\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}

ii) To find y when x=2, substitute:

\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}

\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}

\displaystyle y=\frac{16-72}{16- 3}

\displaystyle y=\frac{-56}{13}

\mathbf{\displaystyle y=-\frac{56}{13}}

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3 years ago
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