Which values are in the solution set of the inequality −2/3x + 13 ≥ −1 ?−23x + 13 ≥ −1 ? Select all that apply. (3 points) Responses 19 19 20 20 21 21 22 22 23
To solve the inequality, we need to isolate the variable x.
−2/3x + 13 ≥ −1
Subtract 13 from both sides:
−2/3x ≥ −1 - 13
−2/3x ≥ -14
To get rid of the fraction, multiply both sides by -3/2 (reciprocal of -2/3):
(-2/3x)(-3/2) ≤ -14(-3/2)
x ≤ 21
Therefore, the solution set is the set of values that are less than or equal to 21. Therefore, the values in the solution set are 19, 20, and 21.
The above compound inequality is true only if _________ of the statement(s) is/are true.(1 point) Responses none none all all at least 1
The above compound inequality is true only if at least 1 of the statement(s) is/are true.
Solve for x. 18 > 2x + 4 ≥ 4 (1 point) Responses 14 < x ≤ 4 14 < x ≤ 4 7 > x > 8 7 > x > 8 7 > x ≥ 12 7 > x ≥ 12 7 > x ≥ 4 7 > x ≥ 4 7 > x ≥ 0
To solve the inequality, we need to isolate the variable x.
Given: 18 > 2x + 4 ≥ 4
Subtract 4 from all sides of the inequality:
14 > 2x ≥ 0
Next, divide all parts of the inequality by 2:
7 > x ≥ 0
So, the solution for x is 7>x≥0.
For which compound inequalities is 6 a solution? Select the statements that are true.(3 points) Responses -5 < x < 6 -5 < x < 6 -2 ≤ x ≤ 6 -2 ≤ x ≤ 6 x ≤ 2 or x ≥ 6 x ≤ 2 or x ≥ 6 x < 2 or x > 6 x < 2 or x > 6 x < -5 or x > 0
To determine if 6 is a solution for each compound inequality, we can substitute 6 into each inequality and see if the statement is true.
-5 < x < 6:
6 does not satisfy this inequality, so it is not a solution.
-2 ≤ x ≤ 6:
6 satisfies this inequality since it is less than or equal to 6, so it is a solution.
x ≤ 2 or x ≥ 6:
6 does not satisfy this inequality, so it is not a solution.
x < 2 or x > 6:
6 does not satisfy this inequality, so it is not a solution.
x < -5 or x > 0:
6 does not satisfy this inequality, so it is not a solution.
Therefore, the only true statement is -2 ≤ x ≤ 6.
A student scored 85 in her Algebra class before she took the End of Course Exam (the EOC). The student wants her average to be between 80 and 90 inclusive after her EOC is entered into her grades. The EOC counts 1/5 of her overall grade and her class average counts 4/5 of her grade. Write and solve a compound inequality to find the possible score she will need to make on the EOC to get the average she wants for her final grade in the course.(1 point) Responses 80 ≤85 + 85 +85 +85 + x5 ≤ 9080 ≤85 + 85 +85 +85 + x5 ≤ 90; 85 ≤ x ≤ 11085 ≤ x ≤ 11080 ≤85 + 85 +85 +85 + x5 ≤ 9080 ≤85 + 85 +85 +85 + x5 ≤ 90; 85 ≤ x ≤ 11085 ≤ x ≤ 110 80 <85 + 85 +85 +85 + x5 < 9080 <85 + 85 +85 +85 + x5 < 90; 85 < x < 11085 < x < 110 80 <85 + 85 +85 +85 + x5 < 9080 <85 + 85 +85 +85 + x5 < 90; 85 < x < 11085 < x < 110 90 ≤85 + 85 +85 +85 + x5 ≤ 8090 ≤85 + 85 +85 +85 + x5 ≤ 80; 110 ≤ x ≤ 80110 ≤ x ≤ 80 90 ≤85 + 85 +85 +85 + x5 ≤ 8090 ≤85 + 85 +85 +85 + x5 ≤ 80; 110 ≤ x ≤ 80110 ≤ x ≤ 80 90 <85 + 85 +85 +85 + x5 < 8090 <85 + 85 +85 +85 + x5 < 80; 110 < x < 85
Let x be the score she needs to make on the EOC.
Her overall grade will be determined by the formula:
(4/5)(class average) + (1/5)(EOC score)
Since she wants her average to be between 80 and 90 inclusive, we can write the compound inequality:
80 ≤ (4/5)*85 + (1/5)*x ≤ 90
Simplifying the inequality:
80 ≤ (340/5) + (1/5)*x ≤ 90
80 ≤ 68 + (1/5)*x ≤ 90
80 - 68 ≤ (1/5)*x ≤ 90 - 68
12 ≤ (1/5)*x ≤ 22
To isolate x, we multiply all parts of the inequality by 5:
5*12 ≤ 5*((1/5)*x) ≤ 5*22
60 ≤ x ≤ 110
Therefore, she will need to score between 60 and 110 on the EOC to get the average she wants for her final grade.