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mash [69]
3 years ago
15

MATTHEW HAS A BAG OF 100 PIECES OF CANDY. HE NEEDS TO SPLIT THE BAG BETWEEN HIS CLASSMATES IN HIS AM AND PM CLASS. HE HAS 20 STU

DENTS IN EACH CLASS. EACH STUDENT WILL NO MORE THAN 4 PIECES OF CANDY AND NO LESS THAN 1 PIECES OF CANDY. HOW MANY PIECES OF CANDY COULD EACH STUDENT POSSIBLY RECEIVE?
Mathematics
1 answer:
DiKsa [7]3 years ago
4 0

Answer: They can recieve about 2.

Exact answer: 2.5

Step-by-step explanation:

If you divded 100 by 40, you get 2.5

Since there are 20 students in each AM and PM classes. you add 20 and 20, which is 40, then you divided by 100 which is 2.5

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I will give brainiest to whoever answers correctly !!
uysha [10]
I don’t know if this correct or not but I think it’s 1,666
8 0
3 years ago
During a sale, a store offered a 15% discount on a TV that originally sold for $420. After the sale, the discounted price of the
wolverine [178]

Answer:

Step-by-step explanation:

Your answer

303.45

Mark it as Brainlist. Follow me for more answer.

7 0
3 years ago
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if the line AB with A(-1,5) and B(3,7) is perpendicular to the line CD WITH C(7,11) AND D(X,23). THEN X=?​
MissTica

Answer:

x=1

Step-by-step explanation:

slope AB = (7-5) / (3 - -1) = 2 / 4 = 1/2

slope CD = (23-11) / (x-7) = 12 / (x-7)  

12 / (x-7) = - 2  ... perpendicular to AB slope 1 = - 1/slope 2

-2 (x-7) = 12

-2x + 14 = 12

-2x = -2

x = 1

5 0
2 years ago
Suppose a triangle has sides a b and c and the angle opposite the side of length a is obtuse.What must be true?
VikaD [51]
When a triangle is obtuse, the length of c^2 is bigger than the combined lengths of a^2 and b^2.

a^2+b^2<c^2

"A" is the best answer.

I hope this helps!
~kaikers
4 0
3 years ago
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James is hosting a game show where a contestant has to draw a ball that is one of four colors from a pot. Each day, a ball of a
Rashid [163]
Let the number of red, black, green and blue balls be R, B, G, U respectively.

B=R  (There are as many black balls as red balls)

G+R=10 so G=10-R (Green balls and red balls should add up to 10)

R+10=U (There should be 10 more blue balls than red balls)

The minimal number of balls is 2, 304 so we have the following inequality:

R+B+G+U≥ 2,304

R+(R)+(10-R)+ (R+10)≥2,304

4R≥2,304

R≥2,304/4=576 so R=576


Answer: 576 
4 0
3 years ago
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