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Setler79 [48]
3 years ago
9

Which expression is equivalent to the following expression? 4.8 (2-3x) + 1.2x​

Mathematics
1 answer:
romanna [79]3 years ago
8 0

Answer:

−13.2x+9.6

Step-by-step explanation:

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Quadrant Four (IV) since the x coordinate is a positive and the y coordinate is a negative.
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Ratios equivalent to 16:12
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\frac{16}{12} equivalent to \frac{ 4}{3}

Step-by-step explanation:

Given ratio, \frac{16}{12}

It can be written as \frac{16}{12}=\frac{4 \times 4}{4 \times 3}

cancel common factor,

\frac{16}{12}=\frac{ 4}{3}

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(J•2)-45=
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C. 10x^2    

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3 years ago
Pioneer company has given an aptitude test to 71 potential job applicants. The test score had an average of 81 and a standard de
Agata [3.3K]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

In point A:

\to df = n - 1 = 71-1= 70

In point B:

\to t = \frac{(x -X)}{s} = \frac{(93-81)}{8} = \frac{12}{8}= 1.5

In point C:

For df = 70, the top 5% critical t score

tcrit = 1.666914479

Thus,

\to 1.666914479 = \frac{(x - 81)}{8}\\\\\to 1.666914479 \times 8 = (x - 81)\\\\\to  13.335315832 = (x - 81)\\\\\to  13.335315832 +81 =x \\\\\to x= 94.335315832

In point D:

For df = 70, the top 5% critical t score

tcrit = -1.666914479

\to -1.666914479 = \frac{(x - 81)}{8}\\\\\to -1.666914479 \times 8= (x - 81)\\\\\to -13.335315832= (x - 81)\\\\\to -13.335315832+81 = x \\\\\to x = 67.664684168\\\\

In point E:

The lower cutoff is 0.10 in the center, which would be around 80 %. The critical point therefore is

tcrit = -1.293762898

\to -1.293762898 = \frac{(x-81)}{8}\\\\\to -1.293762898 \times 8= (x-81) \\\\\to -1.293762898 \times 8= (x-81) \\\\\to -10.350103184=x-81\\\\\to -10.350103184+81=x\\\\\to x=70.649896816

In point F:

The lower cutoff is 0.90 in the center, which would be around 80 %. The critical point therefore is

tcrit = 1.293762898

\to 1.293762898 = \frac{(x-81)}{8}\\\\\to 1.293762898 \times 8= x-81\\\\\to 10.350103184 =x-81\\\\\to 10.350103184 +81=x\\\\\to x = 91.350103184\\

5 0
3 years ago
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