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blagie [28]
3 years ago
12

Mr. and Mrs. Suralbo earned a net taxable income of P568,986. Find the income tax due.

Mathematics
1 answer:
Contact [7]3 years ago
8 0

Answer:

$198,000

Step-by-step explanation:

Since Mr. and Mrs. Suralbo are married and filing jointly, they would fall into the tax slab of 35% as their taxable income ranges between $414,701 to $622,050.

Taxable income = $568,986

Tax rate = 35%

Income tax due = $568,000 * 35/100

= $198,000

Thus, the income tax due for Mr. and Mrs. Suralbo would be $198,000.

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andre [41]
X < -3 is what the answer is
7 0
3 years ago
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Reverse the two digits of my age, divide by three, add 20, and the result is my age. How many years old am I
lapo4ka [179]

Answer:

  • 48 years old.

Step-by-step explanation:

Let the age be xy or 10x + y.

Reverse the two digits of my age, divide by three, add 20, and the result is my age, convert this to equation:

  • (10y + x)/3 + 20 = 10x + y
  • (10y + x)/3 = 10x + y - 20
  • 10y + x = 3(10x + y - 20)
  • 10y + x = 30x + 3y - 60
  • 30x - x + 3y - 10y = 60
  • 29x - 7y = 60

We should consider both x and y are between 1 and 9 since both the age and its reverse are 2-digit numbers.

Possible options for x are:

  • 29x ≥ 7*1 + 60 = 67 ⇒ x > 2, at minimum value of y,

and

  • 29x ≤ 7*9 + 60 = 123 ⇒ x < 5, at maximum value of y.

So x can be 3 or 4.

<h3>If x = 3</h3>

  • 29*3 - 7y = 60
  • 87 - 7y = 60
  • 7y = 27
  • y = 27/7, discarded as fraction.

<h3>If x = 4</h3>

  • 29*4 - 7y = 60
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So the age is 48.

6 0
1 year ago
Student performance across discussion sections: A professor who teaches a large introductory statistics class (197 students) wit
adell [148]

Answer:

The test is not significant at 5% level of significance, hence we conclude that there's no variation among the discussion sections.

Step-by-step explanation:

Assumptions:

1. The sampling from the different discussion sections was independent and random.

2. The populations are normal with means and constant variance

H_0: There's no variation among the discussion sections

H_1: There's variation among the discussion sections

\alpha =0.05

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Section     7       525.01         75             1.87            0.99986

Residuals  189    7584.11        40.13

Test Statistic = F= \frac{75}{40.13} =1.87

Pr(>F) = 0.99986

Since our p-value is greater than our level of significance (0.05), we do not reject the null hypothesis and conclude that there's no significant variation among the eight discussion sections.

3 0
3 years ago
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Answer:

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\sec \theta = \frac{1}{\cos \theta}

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then;

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Substitute the given values we have;

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\tan \theta = \frac{\sin \theta}{\cos \theta}

Substitute the given values we have;

\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}

Therefore, the exact value of:

(a)

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