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RUDIKE [14]
3 years ago
11

Adam tried to compute the average of his 7 test scores. He mistakenly divided the correct sum of all of his test scores by 6, wh

ich yielded 84. What is Adam’s correct average test score?
Mathematics
1 answer:
Harman [31]3 years ago
3 0
Adam got the sum of his grades right, but he divided that correct figure by 6, getting 84 Before that division, the sum of his grades must have been 84 * 6, or 504. He should have divided 504 by 7, getting 72. Answer: His correct average test score is 72.

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12.6 pounds equals how many ounces
Stells [14]
201.6 ounces
Hope this helps and please mark me as brainlest and like
4 0
3 years ago
**Giving away 20 points**. I need to know the arc length for EF and the area of sector EOF please:) (also plz show work thanks!!
LiRa [457]
Let us for a few seconds nevermind there's a circle at all, so we only really have a triangle by itself.

now, EF = 18, wait a second!  EF is the base of the triangle, and h = 10.725, wait a minute!!  "h" is the height of the triangle.

so, what's the area of a triangle whose base is 18 and has a height of 10.725?  yeap, we knew you'd know that one.

what's the length of the arcEF?  well, we know the central angle of ∡FOE is 80°, well, arc's get their angle measurement from the central angle they're in, so if ∡FOE is 80°, so is arcEF then.
3 0
3 years ago
A bag contain 3 black balls and 2 white balls.
Troyanec [42]

Answer:

Step-by-step explanation:

Total number of balls = 3 + 2 = 5

1)

a)

Probability \ of \ taking \ 2 \ black \ ball \ with \ replacement\\\\ = \frac{3C_1}{5C_1} \times \frac{3C_1}{5C_1} =\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\\\\

b)

Probability \ of \ one \ black \ and \ one\ white \ with \ replacement \\\\= \frac{3C_1}{5C_1} \times \frac{2C_1}{5C_1} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}

c)

Probability of at least one black( means BB or BW or WB)

 =\frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5} \\\\= \frac{9}{25} + \frac{6}{25} + \frac{6}{25}\\\\= \frac{21}{25}

d)

Probability of at most one black ( means WW or WB or BW)

=\frac{2}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{2}{5} \times \frac{2}{5} + \frac{3}{5}\\\\= \frac{4}{25} + \frac{6}{25} + \frac{6}{25}\\\\=\frac{16}{25}

2)

a) Probability both black without replacement

  =\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}

b) Probability  of one black and one white

 =\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}

c) Probability of at least one black ( BB or BW or WB)

 =\frac{3}{5} \times \frac{2}{4} + \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{6}{20} + \frac{6}{20} \\\\=\frac{18}{20} \\\\=\frac{9}{10}

d) Probability of at most one black ( BW or WW or WB)

 =\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{2}{20} + \frac{6}{20} \\\\=\frac{14}{20}\\\\=\frac{7}{10}

6 0
3 years ago
Please answer all 3 questions
blsea [12.9K]

Answer:

  • x = 30°
  • DB = 26
  • AD = BC = AB = DC = 7

Step-by-step explanation:

  • <em>Diagonals of a square are congruent and perpendicular and bisect each other</em>
<h3>Q4</h3>

m∠AEB = 3x

m∠AEB = 90°

  • 3x = 90° ⇒ x = 30°
<h3>Q5</h3>

AE = 3x - 2

EC = 2x + 3

  • AE = EC
  • 3x - 2 = 2x + 3
  • 3x - 2x = 3 + 2
  • x = 5

DB = EC = 2(AE) = 2(3*5 - 2) = 2(13) = 26

<h3>Q6</h3>

<u>AD and BC are the sides, which are equal</u>

  • 2x - 1 = 5x - 13
  • 5x - 2x = 13 - 1
  • 3x = 12
  • x = 4

AD = BC = AB = DC =  2*4 - 1 = 7

7 0
2 years ago
Read 2 more answers
Solve for m:3m\5+a=b
Shalnov [3]

Step-by-step explanation:

Assuming:

\frac{3m}{5}  + a = b

subtract a from both sides

\frac{3m}{5}  = b - a

multiply both sides by 5

3m = 5(b - a)

divide both sides by 3

m =  \frac{5(b - a)}{3}

____________________

if,

\frac{3m}{5 + a}  = b

multiply both sides by (5+a), then divide both sides by 3

m =  \frac{b(5 + a)}{3}

3 0
3 years ago
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