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kifflom [539]
3 years ago
8

Marco drew 24 he drew 1/6 of them in art class how many pictures did Marco draw in class

Mathematics
1 answer:
anzhelika [568]3 years ago
5 0
24×1/6 =24/6 = 4

Marco drew 4 pictures
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Solve for x in simplest form 15 = 1/3(x+9)
Lorico [155]

Heya!

Your question states: Solve for x in simplest form 15 = 1/3(x+9)

Answer: X = 36

<em>Numerical Explanation:</em>

1. Step by step

   I. 15 = 1/3(x+9)

2. Simplify both sides of the equation:

    II. 15 = 1/3x + 3

3. Flip the equation:

      III. 1/3x + 3 = 15

4. Subtract 3 from both sides:

          IV. 1/3 x +3 = 15

                          -3    -3

5.  Multiply both sides by 3:

              V.  3 * (1/3 x) = (3) * (12)

6. Final answer:

                  VI. x = 36


I Hoped I Helped!


~KINGJUPITER

         


6 0
3 years ago
Read 2 more answers
In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0
3 years ago
If the check after the substitution of the answer doesn't work, does that mean that the result is incorrect for equation with va
Alecsey [184]
I don't understand what your point is. Mind posting a picture or something?
8 0
3 years ago
What is the slope of the line graph in this picture?
MrMuchimi

Answer:

-5/4

Step-by-step explanation:

y2 - y1

x2 - x1

________

-2 - 3 = -5

5 - 1 = 4

Slope = -5/4

7 0
3 years ago
Una caja tiene 60 bombones eba se comio 1/5 bombones y ana 1/2 bombones cuantos bombones se comio eba y ana
DedPeter [7]

Responder:

42 bombones

Explicación paso a paso:

Dado que :

Número de bombones = 60

Cantidad consumida por Eba = 1/5 * 60 = 12

Cantidad comida por Ana = 1/2 * 60 = 30

Cantidad de chocolate que comieron Eba y Ana:

(12 + 30) = 42

De ahí que Ana y Eba se comieran 42 bombones

7 0
3 years ago
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