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Neporo4naja [7]
3 years ago
14

Jose’s school has 426 students. His principal has promised the Student Council that their idea will be carried out if they can g

et at least 25% of the student population to sign a petition. So far, 82 students have signed the petition. Jose used the following steps to write an inequality that can be used to determine the number of student signatures still needed: Step 1. Declare the variable: Let x = the number of student signatures still needed. Step 2. Create a ratio equivalent to StartFraction total number of signatures needed over total number of students in the school EndFraction : StartFraction x + 82 over 426 EndFraction. Step 3. Convert 25% to a decimal: 25% = 0.25. Step 4. Write the inequality: StartFraction x + 82 over 426 EndFraction less-than-or-equal-to 0.25. What is Jose’s error? In Step 1, x should be equal to the total number of students in the school. In Step 2, the ratio should be StartFraction x over 426 EndFraction. In Step 3, the decimal should be 0.025. In Step 4, the inequality should
Mathematics
2 answers:
Kryger [21]3 years ago
6 0

Answer: Step 4

<u>Step-by-step explanation:</u>

1. Define variable x as student signatures still needed  (Correct)

2)\ \dfrac{x+82}{426}              (Correct)

3. 25% = 0.25      (Correct)

4)\ \dfrac{x+82}{426}\leq 0.25      (Error)!

The inequality symbol should be GREATER THAN or equal to (≥) because Jose needs 25% or more (not less).

Snowcat [4.5K]3 years ago
4 0

Answer:

step 4

Step-by-step explanation:

just took the test and got it right

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4. Diego is thinking of two positive numbers. He says, "If we triple the first number and
aleksandr82 [10.1K]

Answer:

3<em>x </em>+ 2<em>y</em> = 34. and two possible pairs of positive numbers are  (<em>x</em>, <em>y</em>) = (10, 2) and (<em>x</em>, <em>y</em>) = (4, 11).

Step-by-step explanation:

 Let the First positive number be <em>x</em> and second positive number be <em>y.</em>

Triple of first number = 3<em>x</em>

double of second number = 2<em>y</em>

According to question,

3<em>x </em>+ 2<em>y</em> = 34

Therefore, the equation is 3<em>x </em>+ 2<em>y</em> = 34

So the two possible pair of numbers Diego would be thinking of must satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34

Now,  3<em>x </em>+ 2<em>y</em> = 34

3<em>x </em>= 34 - 2<em>y</em>

Let x = 10 and by substituting its value in above expression,

3 \times 10 = 34 - 2y

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2y = 34 - 30

2y = 4

y = 2

Therefore first pair (<em>x</em>, <em>y</em>) = (10, 2)

In the same way put x = 4 then,

3<em>x </em>= 34 - 2<em>y</em>

3 \times 4 = 34 - 2y

2y = 34 - 12

2y = 22

y = 11

Therefore first pair (<em>x</em>, <em>y</em>) = (4, 11)

Therefore, (<em>x</em>, <em>y</em>) = (4, 11) and  (<em>x</em>, <em>y</em>) = (10, 2) are the two possible pairs of numbers Diego could be thinking of as these both values satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34.  

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