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viva [34]
2 years ago
13

Thomas earns a higher pay rate on the number of hours more than 8 that

Mathematics
1 answer:
Yuki888 [10]2 years ago
6 0

The additional percent of his regular pay rate is 82.5% which he earned in the longer shift.

<h3>What is the percentage?</h3>

It is defined as the ratio of two numbers expressed in the fraction of 100 parts. It is the measure to compare two data, the % sign is used to express the percentage.

The longer shift is 12 hours.

Basic pay = $136.00

On a longer shift, Thomas earned = $248.20

The percent in increase:

\rm Percent \ increase = \dfrac{final \ amount - initial \ amount}{initial \ amount}\times 100

\rm Percent \ increase = \dfrac{248.20-136}{136}\times 100

Percent increase = 82.5%

Thus, the additional percent of his regular pay rate is 82.5% which he earned in the longer shift.

Learn more about the percentage here:

brainly.com/question/8011401

#SPJ1

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-5 would Be your answer

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2 years ago
What is 2x - y = last question
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Answer:

Step-by-step explanation:

Let's simplify step-by-step.

2x−y

There are no like terms.

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8 0
3 years ago
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Select the correct answer.
gizmo_the_mogwai [7]

Answer:

Factors are (x-1), (x+1), and (x+2).

Step-by-step explanation:

g(x)=x^{3}+2x^{2} -x-2\\g(x)=(x - 1) (x + 1) (x + 2)

7 0
3 years ago
A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021
egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

  • A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

  • \displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

  • \displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}

Therefore, for the last term we have;

  • \displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

\displaystyle \frac{A}{2}  = \mathbf{ 1011 \cdot  \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460}  \right)}

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2} +\frac{1}{2} -  \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022}  \right)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

  • A ≈ <u>2020.022</u>

Learn more about the sum of a series here:

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Ann [662]

Hey!

so for the first one its 60°, and for the second one its 1/6

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