Ok so 4*5=20
so the area is 20meters
now lets convert 200cm to m
= 200/100=2m
now 20m/2=10
so therefore, the gardener would need 10 bags of grass seeds
hope this helped :)
Answer:
graph 1 = greater than 1
graph 2 = 0
graph 3 = 0
graph 4 = less than 0
graph 5 - between 0 & 1
Sorry if anything is wrong!!
Answer:
The answer is 62
Step-by-step explanation:
Sorry, I meant to add here. You do 98-36 and you get 62.
Answer:
x^(5/6) + 4(x^(7/3))
Step-by-step explanation:
Simplify x to the 1/3 power MULTIPLIED BY (x to the 1/2 power + 2x to the 2 power )
Simplify x^(1/3) × (x^(1/2) + (2x)^2)
= x^(1/3)(x^(1/2)) + x^(1/3)((2x)^2)
= x^(1/3+1/2) + 4(x^(1/3+2))
= x^(5/6) + 4(x^(7/3))
x^(1/3) is y such that y^3 = x
(x^(1/3) × x^(1/3) × x^(1/3)) = x^(1/3+1/3+1/3) = x^1 = x
x^(1/2) = √2 = y such that y^2 = x
(2x)^2 = 4x^2
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
