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ad-work [718]
3 years ago
6

12 divided by 18/25 is?!?!

Mathematics
2 answers:
Dominik [7]3 years ago
6 0

Answer:

50/3 If you convert it to a decimal it's 16.6 if you didn't want an improper fraction it's 16 2/3

alexgriva [62]3 years ago
5 0

Answer:

16.6

Step-by-step explanation:

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All freshmen, sophomores, juniors, and seniors attended a
Trava [24]

Answer:

The probability is .034375 or 11/320

Step-by-step explanation:

the probability of a senior being chose first is 22 (seniors) divided my total number of students (80) is .275 percent.

multiplied by the probability of a sophomore being chosen 10/80 which equals .125.

multiply those together and you get your probability. 0.034375 or 11/320

5 0
2 years ago
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DaniilM [7]
3(x + 7) = 9(x - 1)
3(x) + 3(7) = 9(x) - 9(1)
  3x + 21 = 9x - 9
<u>- 3x          - 3x     </u>
          21 = 3x - 9
        <u>+ 9         + 9</u>
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          3      3
         10 = x
4 0
2 years ago
Help me please this is due today
avanturin [10]

Answer:

490

Step-by-step explanation:

19.6

25

980

3920

4900

then add the decimal

490.0

6 0
3 years ago
Use inductive reasoning to describe the pattern of each sequence. Then find the next two terms. 1, 2, 5, 6, 9,
juin [17]
It looks like the first 1 is the first problem, problem #1. (because of the period)
If this is the case, then the second sequence is 1, 0, -1, ...
1-1=0
0-1=-1
-1-1=-2
-2-1=-3
-3-1=-4
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5 0
3 years ago
Read 2 more answers
A fair 6-faced die is tossed twice. Letting E and F represent the outcomes of the each toss (which are independent), compute the
andreyandreev [35.5K]

Answer:

(a)  \frac{1}{18}            (d)  \frac{1}{9}

(b) \frac{1}{2}               (e) \frac{1}{3}

(c) \frac{4}{9}               (f)  \frac{1}{12}

Step-by-step explanation:

On tossing a 6-face die twice the outcomes of E and F are:

{1, 2, 3, 4, 5 and 6}

And there are total 36 outcomes of the form (E, F).

(a)

The sample space of getting a sum of 11 is: {(5, 6) and (6, 5)}

The probability  of getting a sum of 11 is:

P(Sum 11) =\frac{2}{36} \\=\frac{1}{18}

(b)

The sample space of getting an even sum is:

{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6) (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6) (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}

The probability  of getting an even sum is:

P(Even Sum)=\frac{18}{36}\\ =\frac{1}{2}

(c)

The sample space of getting an odd sum more than 3 is:

{(1, 4), (1, 6), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4),     (5, 6), (6, 1), (6, 3) and (6, 5)}

The probability of getting an odd sum more than 3 is:

<em>P</em> (Odd Sum more than 3) =

=\frac{16}{36}\\=\frac{4}{9}

(d)

The sample space of (E, F) such that E is even and less than 6, and F is odd and greater than 1 is:

{(2, 3), (2, 5), (4, 3) and (4, 5)}

The probability such that E is even and less than 6, and F is odd and greater than 1 is:

P(Even E1) =\frac{4}{36} \\=\frac{1}{9}

(e)

The sample space of E and F such that E is more than 2, and F is less than 4 is:

E = {3, 4, 5 and 6}     F = {1, 2 and 3}

Then the total outcomes of (E, F) will be 12.

The probability such that E is more than 2, and F is less than 4 is:

P(E>2, F

(f)

The sample space of (E, F) such that E is 4 and sum of E and F is odd is:

{(4, 1), (4, 3) and (4, 5)}

The probability such that E is 4 and sum of E and F is odd is:

P(E=4, E+F = Odd)=\frac{3}{36} \\=\frac{1}{12}

6 0
3 years ago
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