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grigory [225]
3 years ago
15

A gift basket that contains jars of jam and packages of bread mix costs $45. there are 8 items in the basket. jars of jam cost $

6 each, and packages of bread mix cost $5 each. write and solve a system of linear equations to find the number of jars of jam and the number of packages of bread mix in the gift basket. let x represent the number of jars of jam and let y represent the number of packages of bread mix.system of equations:
Mathematics
1 answer:
poizon [28]3 years ago
7 0
------------------------------------------------------
Define jam and bread mix
------------------------------------------------------
Let the number of jar of jam be x
Let the number of packages of bread mix be y

------------------------------------------------------
Form equations and solve
------------------------------------------------------
x + y = 8 --------------------- (1)
6x + 5y = 45 ---------------- (2)

------------------------------------------------------
Rewrite equation to make x the subject
------------------------------------------------------
x+ y = 8
x = 8 - y ----------------------------(1a)

------------------------------------------------------
Substitute (1a) into (2)
------------------------------------------------------
6(8 - y) + 5y = 45
48 - 6y + 5y = 45
y = 48 - 45
y = 3

------------------------------------------------------
Substitute y = 3 into (1) to find x
------------------------------------------------------
x + 3 = 8
x  = 8 - 3
x = 5

------------------------------------------------------
Find jam and bread mix
------------------------------------------------------
Jars of Jam = x = 5
Packages of bread mix = y = 3

------------------------------------------------------
Answer: Jam = 5 ; bread mix = 3
------------------------------------------------------
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All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be
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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

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(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.

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