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omeli [17]
3 years ago
8

. Hernandez bought 18 pens for her class. Highlighters cost $3 each, and gel pens cost $2.50 each. She spent a total of $50. Use

a system of equations to find the number of highlighters and gel pens Mrs. Hernandez bought. Enter your answers in the boxes.
Mathematics
2 answers:
cestrela7 [59]3 years ago
5 0
Before we solve for anything, let's assign variables for highlighters and gel pens. Highlighters can be 'h,' and gel pens can be 'g'.

We can make two equations from what we've been given so far. Since we know that Hernandez bought 18 pens, we know that:

h + g = 18

And since we know the cost of an individual gel pen and highlighter as well as the total price, we can make another equation:

3h + 2.50g = 50

We can simplify the first equation by keeping only one variable on one side:

h + g = 18
h = 18 - g

Now that we have a value for h, we can assign it to the second equation:

3h + 2.50g = 50
3(18 - g) + 2.50g = 50
54 - 3g + 2.50g = 50

Simplify:
-0.50g = -4
0.5g = 4
5g = 40
g = 8

Put this value into the equation we made earlier:

h = 18 - g
h = 18 - 8
h = 10

Hernandez bought 10 highlighters and 8 gel pens.
Zolol [24]3 years ago
4 0

Answer:

There are 9 highlighters and  8 gel pens

Step-by-step explanation:

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4 0
3 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
H/5 = 5.1<br> What is H?
inysia [295]
This basically says all the steps ! If you have any questions just tell me :)

5 0
3 years ago
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Let us assume the number of turtles = x
Then
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x + 2x + 3 = 54
3x + 3 = 54
3x = 54 - 3
3x = 51
x = 51/3
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3 years ago
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