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ella [17]
3 years ago
6

Kevin picked 3/4 of a pound of strawberries. That afternoon, his sister ate 2/3 of the strawberries Kevin picked. How many pound

s of strawberries did Kevin's sister eat? Use a visual model (such a number line, area diagram or pattern blocks) to find the answer to the problem. State the answer and explain how you determined the answer from your model.
Mathematics
1 answer:
Lunna [17]3 years ago
5 0

Answer:

1/12 pounds of strawbebbies left.

Step-by-step explanation:

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The manager of a snack bar buys bottled water in packs of 35 and candy bars in packs of 20. Then, she sells the items individual
Zolol [24]
If you do 35 times 4 you get 140. Divide 140 by 20 and you get 7 . Therefore the manager would have to buy 7 packs of candy bars and 4 packs of water bottles
5 0
3 years ago
Find the location of Y, the midpoint of XZ X is located at (3,-9)
Levart [38]

The location of point Y is the point -3 on the number line

Here, we want to find the location of Y

let us say the point 0 represents the origin

The coordinates of point x will be (-9,0) while the coordinates of point z will be (3,0)

To find the location of Y , we proceed to use the midpoint formula as follows;

(x,y)=\frac{x_{1\text{ }}+x_2}{2},\frac{y_{1\text{ }}+y_2}{2}

Thus, we have;

(-9 + 3)/2 , (0 + 0)/2

= -6/2 , 0

= (-3 , 0)

3 0
1 year ago
There are 64 students in a speech contest. Yesterday, Ā of them gave their speeches. Today, 7 of the remaining students gave the
Alexxx [7]

Answer:

23

Step-by-step explanation:

Ā means the average

So it means the average of 64 gave their speech which should be 64/2 = 32

So 32 gave their speeches yesterday,

7 gave today

Total number that has given is

32 + 7 = 39

Therefore the number remaining are

62 - 39 = 23

8 0
3 years ago
A different class has 285 students, and 51.2% of them are men. how many men are in the class?​
Zanzabum

Answer:

146

Step-by-step explanation:

Get a percent calculator, Then, you put 51.2% of 285 and you actually get 145.96 but the rounded of what you need is 146

6 0
3 years ago
Show that in a group of five people (where any two people are either friends or enemies), there are not necessarily three mutual
sveta [45]

Answer:

Lets call the people A,B,C,D and E. We need to find an example where there are not 3 mutual friends and not three mutual enemies.

Lets start by assuming that A has 2 friends, B and C, and 2 enemies, D and E.

Since we want A,B and C not to be mutually friends, then B and C neccesarily have to be enemies.

Now, since B and C are enemies, they cant be enemies at the same time with D, and they also cant be enemies at the same time with E. Therefore one of them has to be friends with D and one has to be friends with E.

Also, since D and E are enemies with A, they neccesarily need to be friends.

From the fact that D and E are friends, we coclude that they cant have a friend in common, as a consequence, B has to be friends with one of D and E and C has to be friend with the other. We can assume that B is friend with D and C is friend with E. If we use that, we have the following friendship configuration

---------------------

Friends of A: B, C

Enemies of A: D,E

-------------------------

Friends of B: A,D

Enemies of B: C, E

------------------------

Friends of C: A,E

Enemies of C: B,D

------------------------

Friends of D: B,E

Enemies of D: A,C

----------------------------

Friends of E: C,D

Enemies of E:  A,B

--------------------------

As you can check, there is not three mutual friends nor three mutual enemies.

8 0
3 years ago
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