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Svetllana [295]
3 years ago
13

Solve for x. -4.5 (x - 8.9) = 12.6 Enter your answer, as a decimal, in the box.

Mathematics
2 answers:
Ksju [112]3 years ago
7 0
For a Brainliest?

<span>Simplifying 21.5x + -7.7 + -12.6x + 17 = 0 Reorder the terms: -7.7 + 17 + 21.5x + -12.6x = 0 Combine like terms: -7.7 + 17 = 9.3 9.3 + 21.5x + -12.6x = 0 Combine like terms: 21.5x + -12.6x = 8.9x 9.3 + 8.9x = 0 Solving 9.3 + 8.9x = 0 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-9.3' to each side of the equation. 9.3 + -9.3 + 8.9x = 0 + -9.3 Combine like terms: 9.3 + -9.3 = 0.0 0.0 + 8.9x = 0 + -9.3 8.9x = 0 + -9.3 Combine like terms: 0 + -9.3 = -9.3 8.9x = -9.3 Divide each side by '8.9'. x = -1.04494382 Simplifying x = -1.04494382</span>
ollegr [7]3 years ago
5 0
I believe that the answer would be 26. First, put 26 for x and subtract 8.9 from 26 s 26 - 8.9 = 17.1. Next, you will subtract -4.5 from 17.1 which equals 12.6. So 26 is your answer.
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Answer:

the student should score atleast 229 to be among the top 10%.

Step-by-step explanation:

in terms of the normal distribution, and if the table that you're using calculates the area of the normal distribution from the mean to a point x, only then what we are actually finding the value 'x' at which the z-score is at 40% (the rest 50% is already skipped by the table)

P(0.4) = \dfrac{x - \mu}{\sigma}

after finding the the value at this z-score, we can find the value of x at which the score is in the top 10% range.

we can find the z-score either using a normal distribution table or calculator. (but be sure what area is it calculating)

looking at the table the closest value we can find is, 0.4015 at z = 1.29 ((it is above 40% because we want to be in the top 10% range)

P(0.4015) = \dfrac{x - 100}{100}

1.29 = \dfrac{x - 100}{100}

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the student should score atleast 229 to be among the top 10%.

7 0
2 years ago
What fraction is equivalent to 4/24​
matrenka [14]

Answer:

1/6

Step-by-step explanation:

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6 0
3 years ago
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Hope I helped :) 
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