Question 1

Solve the inequality. Graph your solution on a number line, and write your solution in interval notation
-$|8 x+3|-2>0$

A) $ \left(-\frac{5}{8},-\frac{1}{8}\right) $

B) $ \left(-\infty,-\frac{5}{8}\right) 
\cup\left(-\frac{1}{8}, \infty\right) $

C) $ \left(-\frac{1}{8}, \infty\right) $

D) $ (-\infty,-8) \cup(8, \infty) $

Accepted Answer

The answer of Solve the inequality. Graph your solution on...

Question 2

Solve the inequality. Graph your solution on a number line, and write your solution in interval notation
-$6-|x+3| \geq-9$

A) $[-18,12]$

B) $[-15,15]$

C) $(-\infty,-18] \cup[12, \infty)$

D) $(-\infty, 12)$

Accepted Answer

The answer of Solve the inequality. Graph your solution on...

Question 3

Solve the inequality. Graph your solution on a number line, and write your solution in interval notation
-$3-|4-2 \mathrm{x}|>-3$

A) $ (-1,5) $

B) $ (-1, \infty) $

C) $ (-\infty,-1) \cup(5, \infty) $

D) $ (-\infty,-1) $

Accepted Answer

The answer of Solve the inequality. Graph your solution on...

Question 4

Solve the inequality. Graph your solution on a number line, and write your solution in interval notation
-$|8 x-4|-9>-18$

A) $(-\infty, \infty)$
 
B) $\left(-\infty,-\frac{5}{8}\right) \cup\left(\frac{13}{8}, \infty\right)$
 
C) $\left(-\frac{5}{8}, \infty\right)$
 
D) $\varnothing$

Accepted Answer

The answer of Solve the inequality. Graph your solution on...

Question 5

Solve the inequality. Graph your solution on a number line, and write your solution in interval notation
-$|7-5 x| \geq 1$

A) $ \left(-\infty, \frac{6}{5}\right] \cup\left[\frac{8}{5}, \infty\right) $

B) $ \left[-\frac{8}{5},-\frac{6}{5}\right] $

C) $ \left[\frac{6}{5}, \frac{8}{5}\right] $

D) $ \left(-\infty,-\frac{8}{5}\right] \cup\left[-\frac{6}{5}, \infty\right) $

Accepted Answer

The answer of Solve the inequality. Graph your solution on...

Question 6

Construct a 95% confidence interval for the population mean using the given information.
-New medication's effect on total serum cholesterol level of all patients (drop in total cholesterol), $\overline{\mathrm{x}}=10.6 \mathrm{mg} / \mathrm{dL}, \mathrm{E}=3.1 \mathrm{mg} / \mathrm{dL}$
A) $7.6<\mu<13.8$
B) $7.5<\mu<13.7$
C) $7.4<\mu<13.9$
D) $7.6<\mu<13.6$

Accepted Answer

The answer of Construct a 95% confidence interval for the...

Question 7

Construct a 95% confidence interval for the population mean using the given information.
-Starting annual salary for all mechanical engineering graduates, $\overline{\mathrm{x}}=\$ 68,400, \mathrm{E}=\$ 14,280$
A) $\$ 61,260<\mu<\$ 75,540$
B) $\$ 54,120<\mu<\$ 82,680$
C) $\$ 52,140<\mu<\$ 86,620$
D) $\$ 62,210<\mu<\$ 74,550$

Accepted Answer

The confidence interval is calculated as $\overline{\mathrm{x}} \pm \mathrm{E}$, so $68,400 \pm 14,280 = (54,120, 82,680)$.

Question 8

The confidence interval for a population pencentage is generated from the inequality $|\mathbf{p}-\hat{\mathbf{p}}|<\mathrm{E}$, where $\mathrm{E}$ is the margin of error. Construct a $95 \%$ confidence interval for the population percentage using the given information.
-Percentage of false positives generated by a new diabetes screening test, $\hat{\mathrm{p}}=8.5 \%, \mathrm{E}=4.6 \%$
A) $3.7 \%< p<13.2 \%$
B) $4.1 \%< p<13.0 \%$
C) $4.1 \%< p<13.2 \%$
D) $3.9 \%< p<13.1 \%$

Accepted Answer

The answer of The confidence interval for a population pencentage...

Question 9

Classify the absolute value inequality as
· Having a single interval containing its solutions, but not the entire set of real numbers
· Having the union of two disjoint intervals containing its solutions
· Being a contradiction with no solution
· Being an identity with the set of all real numbers as its solution set
-$|3 x+4|+11 \leq 5$
A) identity
B) union of two intervals
C) single interval
D) contradiction

Accepted Answer

The given inequality $|3x + 4| + 11 \leq 5$ simplifies to $|3x + 4| \leq -6$. Since the absolute value of any real number is always non-negative, there are no real numbers $x$ that can satisfy this inequality, making it a contradiction with no solution.

Question 10

Classify the absolute value inequality as
· Having a single interval containing its solutions, but not the entire set of real numbers
· Having the union of two disjoint intervals containing its solutions
· Being a contradiction with no solution
· Being an identity with the set of all real numbers as its solution set
-$|4 \mathrm{x}-1|-8 \leq 3$
A) identity
B) union of two intervals
C) single interval
D) contradiction

Accepted Answer

The answer of Classify the absolute value inequality as
· Having...

Question 11

Classify the absolute value inequality as
· Having a single interval containing its solutions, but not the entire set of real numbers
· Having the union of two disjoint intervals containing its solutions
· Being a contradiction with no solution
· Being an identity with the set of all real numbers as its solution set
-$|x-5|-11 \geq 7$
A) union of two intervals
B) identity
C) single interval
D) contradiction

Accepted Answer

The given inequality $|x-5|-11 \geq 7$ simplifies to $|x-5| \geq 18$, which means the distance between $x$ and 5 is at least 18. This results in two intervals: one to the left of 5 and one to the right, hence the union of two disjoint intervals.

Question 12

Classify the absolute value inequality as
· Having a single interval containing its solutions, but not the entire set of real numbers
· Having the union of two disjoint intervals containing its solutions
· Being a contradiction with no solution
· Being an identity with the set of all real numbers as its solution set
-$|6 x+7|-2>-13$
A) union of two intervals
B) contradiction
C) identity
D) single interval

Accepted Answer

The answer of Classify the absolute value inequality as
· Having...

Question 13

Show that for any positive number $\varepsilon$ we can find a positive number $\delta$ such that the statement below is satisfied.
-$6 \mathrm{x}-7$ is less than $\varepsilon$ away from 5 whenever $\mathrm{x}$ is less than $\delta$ away from 2.
A) $\delta=\frac{\varepsilon}{6}$

B) $\delta=\frac{\varepsilon}{8}$

C) $\delta=\frac{\varepsilon}{4}$

D) $\delta=\frac{\varepsilon}{7}$

Accepted Answer

The given condition translates to $|6x - 7 - 5| < \varepsilon$ when $|x - 2| < \delta$. Simplifying the left side gives $|6x - 12| < \varepsilon$, or $|6(x - 2)| < \varepsilon$, leading to $|x - 2| < \frac{\varepsilon}{6}$, hence $\delta = \frac{\varepsilon}{6}$.

Question 14

Show that for any positive number $\varepsilon$ we can find a positive number $\delta$ such that the statement below is satisfied.
-15x - 125 is less than $\varepsilon$ away from 130 whenever $\mathrm{x}$ is less than $\delta$ away from - 17 .
A) $\delta=\frac{\varepsilon}{-15}$

B) $\delta=\frac{\varepsilon}{-17}$

C) $\delta=\frac{\varepsilon}{17}$

D) $\delta=\frac{\varepsilon}{15}$

Accepted Answer

To satisfy the condition, we need $|15x - 125 - 130| < \varepsilon$, which simplifies to $|15x - 255| < \varepsilon$. This implies $|15(x + 17)| < \varepsilon$, leading to $|x + 17| < \frac{\varepsilon}{15}$. Therefore, $\delta = \frac{\varepsilon}{15}$.

Quiz 2: Linear and Absolute Value Equations and Inequalities

Construct a 95% confidence interval for the population mean using the given information. -New medication's effect on total serum cholesterol level of all patients (drop in total cholesterol) , $\overline{\mathrm{x}}=10.6 \mathrm{mg} / \mathrm{dL}, \mathrm{E}=3.1 \mathrm{mg} / \mathrm{dL}$

Construct a 95% confidence interval for the population mean using the given information. -Starting annual salary for all mechanical engineering graduates, $\overline{\mathrm{x}}=\$ 68,400, \mathrm{E}=\$ 14,280$

Show that for any positive number $\varepsilon$ we can find a positive number $\delta$ such that the statement below is satisfied. - $6 \mathrm{x}-7$ is less than $\varepsilon$ away from 5 whenever $\mathrm{x}$ is less than $\delta$ away from 2.

Show that for any positive number $\varepsilon$ we can find a positive number $\delta$ such that the statement below is satisfied. -15x - 125 is less than $\varepsilon$ away from 130 whenever $\mathrm{x}$ is less than $\delta$ away from - 17 .

Quiz 2: Linear and Absolute Value Equations and Inequalities

Solve the inequality. Graph your solution on a number line, and write your solution in interval notation - ∣8x+3∣−2>0|8 x+3|-2>0∣8x+3∣−2>0

Solve the inequality. Graph your solution on a number line, and write your solution in interval notation - 6−∣x+3∣≥−96-|x+3| \geq-96−∣x+3∣≥−9

Solve the inequality. Graph your solution on a number line, and write your solution in interval notation - 3−∣4−2x∣>−33-|4-2 \mathrm{x}|>-33−∣4−2x∣>−3

Solve the inequality. Graph your solution on a number line, and write your solution in interval notation - ∣8x−4∣−9>−18|8 x-4|-9>-18∣8x−4∣−9>−18

Solve the inequality. Graph your solution on a number line, and write your solution in interval notation - ∣7−5x∣≥1|7-5 x| \geq 1∣7−5x∣≥1

Construct a 95% confidence interval for the population mean using the given information. -Starting annual salary for all mechanical engineering graduates, x‾=$68,400,E=$14,280\overline{\mathrm{x}}=\$ 68,400, \mathrm{E}=\$ 14,280x=$68,400,E=$14,280

Show that for any positive number ε\varepsilonε we can find a positive number δ\deltaδ such that the statement below is satisfied. - 6x−76 \mathrm{x}-76x−7 is less than ε\varepsilonε away from 5 whenever x\mathrm{x}x is less than δ\deltaδ away from 2.

Show that for any positive number ε\varepsilonε we can find a positive number δ\deltaδ such that the statement below is satisfied. -15x - 125 is less than ε\varepsilonε away from 130 whenever x\mathrm{x}x is less than δ\deltaδ away from - 17 .

Construct a 95% confidence interval for the population mean using the given information. -Starting annual salary for all mechanical engineering graduates, $\overline{\mathrm{x}}=\$ 68,400, \mathrm{E}=\$ 14,280$

Show that for any positive number $\varepsilon$ we can find a positive number $\delta$ such that the statement below is satisfied. - $6 \mathrm{x}-7$ is less than $\varepsilon$ away from 5 whenever $\mathrm{x}$ is less than $\delta$ away from 2.

Show that for any positive number $\varepsilon$ we can find a positive number $\delta$ such that the statement below is satisfied. -15x - 125 is less than $\varepsilon$ away from 130 whenever $\mathrm{x}$ is less than $\delta$ away from - 17 .