Answer:
14
Step-by-step explanation:
Omar can make 14 whole dumplings. You change 2 3/4 cups to an improper fraction ( multiply the denominator and add the numerator ) and you get 11/4. Since you need 3/16 cups for one dumpling you have to find the greatest common denominator which is 16. You multiply 4 by 11 to get the numerator and end up with 44/16. Since you need 3/16 for one whole dumping you divide 44 by 3 and get 14.6 repeating. You cannot have a fraction of a dumpling so you round down and get the answer 14.
Answer:
Option D. 8 units
Step-by-step explanation:
step 1
Find the area of rectangle
The area of rectangle is
![A=(4)(16)=64\ units^2](https://tex.z-dn.net/?f=A%3D%284%29%2816%29%3D64%5C%20units%5E2)
step 2
Find the length side of the square with the same area of rectangle
The area of a square is
![A=b^2](https://tex.z-dn.net/?f=A%3Db%5E2)
where
b is the length side of the square
we have
![A=64\ units^2](https://tex.z-dn.net/?f=A%3D64%5C%20units%5E2)
substitute
![64=b^2](https://tex.z-dn.net/?f=64%3Db%5E2)
take the square root both sides
![b=8\ units](https://tex.z-dn.net/?f=b%3D8%5C%20units)
therefore
The length side of the square is 8 units
Answer:
Step-by-step explanation:
Slope of line passing through (2, 7) and (1, -2) = (-2-7)/(1-2) = 9
Answer:
thats great =)
Step-by-step explanation: can i pls have brainliest pls im on one more
1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:
A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?
You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked
2) a - 2% as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.
b - 1,000,000/2500 = 400
400 packages are defective each year