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timurjin [86]
3 years ago
15

A recent survey at a high school found that 3 out of 5 seniors played atleast one sport during high school.

Mathematics
2 answers:
stira [4]3 years ago
8 0
D would show three fifths of seniors played a sport
Lena [83]3 years ago
4 0

Answer:

D, 150 Seniors out of 250

Step-by-step explanation:

Multiply 3*150, and 5*150.

150 over 250 is the same as 3 out of 5.

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Please help me........
kati45 [8]

Answer:

x = 54

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define</u>

-12(x + 4) + 14x = 60

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Distribute -12:                              -12x - 48 + 14x = 60
  2. Combine like terms:                   2x - 48 = 60
  3. Add 48 to both sides:                 2x = 108
  4. Divide 2 on both sides:              x = 54
7 0
3 years ago
Read 2 more answers
Work out<br> 2/21<br> +<br> 6/7<br> Give your answer in its simplest form.
olganol [36]

Answer:

2/21+6/7 = 20/21

Step-by-step explanation:

2/21+6/7

=2/21+(6*3/7*3)

=2/21+18/21

=20/21

Hope this helps! Comment any questions <3

4 0
2 years ago
Read 2 more answers
What is the solution to the given inequality?
xenn [34]

Answer: x ≤3

Step-by-step explanation:

Your answer is correct

7 0
3 years ago
-112/3 x (-4 1/5) =?
TEA [102]

Answer:

sorry Wala po akong answer ehhh sorry

3 0
3 years ago
Read this E[2X^2 â€" Y].
djyliett [7]

Looks like a badly encoded/decoded symbol. It's supposed to be a minus sign, so you're asked to find the expectation of 2<em>X </em>² - <em>Y</em>.

If you don't know how <em>X</em> or <em>Y</em> are distributed, but you know E[<em>X</em> ²] and E[<em>Y</em>], then it's as simple as distributing the expectation over the sum:

E[2<em>X </em>² - <em>Y</em>] = 2 E[<em>X </em>²] - E[<em>Y</em>]

Or, if you're given the expectation and variance of <em>X</em>, you have

Var[<em>X</em>] = E[<em>X</em> ²] - E[<em>X</em>]²

→   E[2<em>X </em>² - <em>Y</em>] = 2 (Var[<em>X</em>] + E[<em>X</em>]²) - E[<em>Y</em>]

Otherwise, you may be given the density function, or joint density, in which case you can determine the expectations by computing an integral or sum.

6 0
3 years ago
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