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3 years ago

Carly can paint 6/7 of a picture in 3/14 of an hour how many pictures can they paint in a full hour

2 answers:

6 0

28/7 which is equivalent to 4 painting in a full hour, this is because Carly can paint 2/7 in 1/14 of an hour multiply the 2 times 14 and get 28/7 or 4 which is you’re answer

Svetradugi [14.3K]3 years ago

6 0

Answer:

4 paintings in a full hour

Work:

Carly painted 6/7 of a picture in 3/14 hours:

6/7=3/14

Which means Carly painted 2/7 of a picture in 1/14 hours:

2/7=1/14

We need to find how many paintings she can paint in 1 full hour, or 14/14 of an hour, how do you get from 1/14 to 14/14? You do multiplication:

1/14 x 14 = 14/14

but there is a rule that says what you do to one side of the equation, you need to do it to the other side too so:

2/7 x 14=1/14 x 14

28/7=14/14

28/7=4

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$(a)P(Red)=\dfrac{2}{3}\\(b)P(Odd) =\dfrac{1}{2}\\(c)P(\text{Red or Odd Numbered})=\dfrac{5}{6}\\(d)P(\text{Blue or Even Numbered})=\dfrac{2}{3}$

Step-by-step explanation:

Number of Red Marbles{1,2,3,4,5,6,7,8,9,10,11,12},n(R)=12

Number of Blue Marbles{1,2,3,4,5,6},n(B)=6

Total Number of Marbles, n(S)=6+12=18

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(b)Probability that the marble is odd-numbered.

Number of Odd-Numbered Balls, n(O)=9

$P(Odd)=\dfrac{n(O)}{n(S)} =\dfrac{9}{18} =\dfrac{1}{2}$

(c)Probability that the marble is red or odd-numbered.

n(Red)=12

n(Odd Numbered marbles)=9

n(Red and Odd Numbered Marbles)=6

$P(\text{Red or Odd Numbered})=P(Red)+P(Odd\:Numbered)-P(\text{Red and Odd Numbered)}\\=\dfrac{12}{18} +\dfrac{9}{18}-\dfrac{6}{18} =\dfrac{15}{18}\\P(\text{Red or Odd Numbered})=\dfrac{5}{6}$ (d)Probability that the marble is blue or even-numbered.

n(Blue)=6

n(Even Numbered marbles)=9

n(Blue and Even Numbered Marbles)=3

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