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hammer [34]
3 years ago
10

A professor grades students on four tests, a term paper, and a final examination. Each test counts as 15% of the course grade. T

he term paper counts as 20% of the course grade. The final examination counts as 20% of the course grade. Alan has test scores of
92, 78, 82, and 90.
Alan received an 80 on his term paper. His final examination score was 86. Use the weighted mean formula to find Alan's average for the course. (Round your answer to one decimal place.)

Mathematics
1 answer:
alina1380 [7]3 years ago
5 0

Answer:

The Alans's average for the course is 84.5

Step-by-step explanation:

We are given

There are 4 tests, A Term paper and a Final examination.

Score = 92, 78, 82, 90.

Term paper score = 80

Final examination score = 86

Weighted mean = ∑ w.f/∑w

     also the sum of all the weights is 100% = 1

weighted mean = 15%*92 + 15%*78+15%*82+15%*90+20%*80+20%*86/1

                         = 51.3 + 33.2

                         =  84.5

Therefore the Alans's average for the course is 84.5

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Find the area of quadrilateral ABCD
andreev551 [17]

Answer:

<em>A ≈ 28.5</em>

Step-by-step explanation:

a, b, c

P = a + b + c

Semiperimeter s = \frac{a+b+c}{2}

A = \sqrt{s(s-a)(s-b)(s-c)}

~~~~~~~~~~~~~~~

P_{ABC} = 4.3 + 2.89 + 6.81 = 14

s = 14 ÷ 2 = 7

A_{ABC} = \sqrt{7(7-4.3)(7-2.89)(7-6.81)} = √14.75901 ≈ 3.84

P_{BCD} = 8.59 + 7.58 + 6.81 = 22.98

s = 22.98 ÷ 2 = 11.49

A_{BCD} = \sqrt{11.49(11.49-8.59)(11.49-7.58)(11.49-6.81)} = √609.7343148 ≈ 24.6928

A_{ABCD} = 3.84 + 24.6928 ≈ <em>28.5</em>

3 0
3 years ago
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Nostrana [21]
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3 0
3 years ago
Which of the following logarithms do not exist?
Rasek [7]

Answer:

B. logarithms of negative number don't exist

5 0
3 years ago
Please help with this question!!
Goshia [24]

Answer:C

Step-by-step explanation:

4 0
3 years ago
Kite EFGH is inscribed in a rectangle where F and H are midpoints of parallel sides. The area of EFGH is 35 square units. What i
sasho [114]

*see attachment for the figure described

Answer:

5 units

Step-by-step explanation:

==>Given the figure attached below, let where FH and EG intercepted be K.

Since FH are midpoints of parallel lines, KE = KG = x.

Given that the area of the kite EFGH = 35 square units, and we know the length of one of the diagonals = HF = KF + KH = 2 + 5 = 7, we can solve for x using the formula for the area of a kite.

Area of kite = ½ × d1 × d2

Where d1 = KH = 7

d2 = EG = KE + KG = x + x = 2x

Area of kite EFGH = 35

THUS:

35 = ½ × 7 × 2x

35 = 1 × 7 × x

35 = 7x

Divide both sides by 7

35/7 = x

x = 5

4 0
3 years ago
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