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KiRa [710]
3 years ago
10

Mark has 23 feet of string. He will cut the string into equal lengths of 9 inches each. How much string will be left after he cu

ts as many pieces of these lengths as possible?
Mathematics
1 answer:
Elza [17]3 years ago
8 0

Convert foot to inches:

1 foot = 12 inches

23 feet ⇒  23 feet × 12 inches/ft = 276 inches

Divide:

276 ÷ 9 = 30 pieces with remainder

Remainder = 276 - (9×30)  ⇒  276 - 270 = 6

ANSWER: There will be 6 inches long remaining in length after cutting the string into 30 pieces equal lenght of 9 inches each.

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Phil has 7 quarts of hot chocolate to give his class mates. how many of phils friend's can have one cup of hot chocolate
Rudik [331]
4 cups per quarter

4 x 7 = 28

28 kids can have a cup of hot chocolate.
4 0
3 years ago
Find an expression which represents the difference when (-3x + 5y) is subtracted
lubasha [3.4K]

Step-by-step explanation:

-3x+5y-(-6x)+3y

-3x+5y+6x+3y

-3x+6x+5y+3y

3x+2y.

8 0
2 years ago
All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be
Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

A = a + (A \cap B)

In which a is the probability that a student is proficient in reading but not mathematics and A \cap B is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

B = b + (A \cap B)

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

(A \cup B) + C = 1

In which

(A \cup B) = a + b + (A \cap B)

65% were found to be proficient in both reading and mathematics.

This means that A \cap B = 0.65

78% were found to be proficient in mathematics

This means that B = 0.78

B = b + (A \cap B)

0.78 = b + 0.65

b = 0.13

85% of the students were found to be proficient in reading

This means that A = 0.85

A = a + (A \cap B)

0.85 = a + 0.65

a = 0.20

Proficient in at least one:

(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98

What is the probability that the student is proficient in neither reading nor mathematics?

(A \cup B) + C = 1

C = 1 - (A \cup B) = 1 - 0.98 = 0.02

There is a 2% probability that the student is proficient in neither reading nor mathematics.

6 0
3 years ago
Question 6 Unsaved Identify the slope and y-intercept of the graph of the equation. Then graph the equation. y = −5/4x+1
Lena [83]
Slope = -5/4
y intercept = 1

The slope is always attached to x. The intercept is the number attached on the end. 
4 0
3 years ago
There are 1,570 souvenir paperweights that need to be packed in boxes. Each box will hold 17 paperweights. How many boxes will b
arlik [135]

Answer:

93 boxes will be needed.

Step-by-step explanation:

Divide 1570 ÷ 17

This gives the answer 92.35

So 92 boxes won't be enough. 0.35 of a box itsn't helpful in the real world. So you need 93 boxes.

8 0
2 years ago
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