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4 years ago

8 friends share 3 cookies, what fraction of the cookie does each friend get?

1 answer:

4 0

8 friends share 3 cookies.

There are 3 cookies and 8 people.

We need to know how much cookie 1 friend gets.

1 friend gets 3/8 of a cookie.

There are 8 friends, so 8 x 3/8 = 3. The 8 friends will finish 3 cookies if they all have 3/8 of a cookie.

Each friend gets 3/8 of a cookie.

You might be interested in

What is the distance between the point (−10, −2)(−10, −2) and the point (−4, −2)(−4, −2) ?

Gnoma [55]

In this case: x 1 = - 10, y 1 = - 2;
x 2 = - 4 , y 2 = - 2
The distance between the points :
$d = \sqrt{(x 2 - x 1 ) ^{2} + ( y 2 - y 1 ) ^{2} } = \\ =\sqrt{(-4+10) ^{2} +(-2+2) ^{2} } = \\ = \sqrt{6 ^{2}+0 ^{2} } = \sqrt{36}= 6$
Answer: d = 6

4 0

3 years ago

Read 2 more answers

A local hamburger shop sold a combined total of 712 hamburgers and cheeseburgers on Tuesday. There were 62 more cheeseburgers so

hammer [34]

A local hamburger shop sold a total of 712 burgers on Tuesday

There are 62 more cheeseburgers than hamburgers

let cheeseburers = c
let hamburgers = h

h + 62 = c
c + h = 712

Plug in h + 62 for c

(h + 62) + h = 712
2h + 62 = 712
2h + 62 (-62) = 712 (-62)
2h = 650
2h/2 = 650/2
h = 325

There are 325 hamburgers sold on Tuesday

c = 325 + 62
c = 387

There are 387 chesseburgers sold on tuesday (in case you were wondering)

hope this helps

4 0

3 years ago

In a survey, 32% of the people chose pizza as their favorite food. The results were reported to within 2% accuracy. What is the

statuscvo [17]

Answer:

okogfyuijsudiwgsydjhdbdjwnwbwv

4 0

3 years ago

a father is now 38 years older than his son. Ten years ago he was twice as old as his son. let x represent the age of the son no

Helga [31]

The present age of father is 86 years old and present age of son is 48 years old

Solution:

Given that, a father is now 38 years older than his son

Ten years ago he was twice as old as his son

Let "x" be the age of son now

Therefore, from given,

Father age now = 38 + age of son now

Father age now = 38 + x

Ten years ago he was twice as old as his son

Age of son ten years ago = age of son now - 10

Age of son ten years ago = x - 10

Age of father ten years ago = 38 + x - 10

Then we get,

Age of father ten years ago = twice the age of son ten years ago

38 + x - 10 = 2(x - 10)

28 + x = 2x - 20

2x - x = 28 + 20

x = 48

Thus son age now is 48 years old

Father age now = x + 38 = 48 + 38 = 86

Thus present age of father is 86 years old and present age of son is 48 years old

7 0

4 years ago

Solve the following inequality algebraically: look at the picture

seraphim [82]

The answer is b -9 < x < 12

8 0

3 years ago