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iragen [17]
4 years ago
12

12x+7<−11 OR 5x−8>40 solve for x

Mathematics
2 answers:
Flura [38]4 years ago
7 0

Answer:

x < -3/2 OR x > 9  3/5

Step-by-step explanation:

12x+7<−11 OR 5x−8>40

Solve the inequalities separately

12x+7<−11

Subtract 7 from each side

12x+7-7<−11 -7

12x < -18

Divide each side by 12

12x/12 < -18/12

x < -3/2

Then solve the other inequality

5x−8>40

Add 8 to each side

5x-8+8>40+8

5x>48

Divide each side by 5

5x/5 >48/5

x > 9  3/5

Put them back together with the OR in the middle

x < -3/2 OR x > 9  3/5

kipiarov [429]4 years ago
6 0

Answer:

x < -1.5 U x > 9.6

Step-by-step explanation:

12x + 7 < -11

x < -1.5

5x - 8 > 40

5x > 48

x > 9.6

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Answer:

t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115    

Now we can calculate the degrees of freedom

df=n-1=15-1=14  

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got t_{critc}= 1.761

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

Step-by-step explanation:

Information given

\bar X=147.3 represent the sample mean for the amount of time spent at part time jobs

s=50 represent the sample standard deviation

n=15 sample size  

\mu_o =120 represent the value to check

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value for the test

System of hypothesis

We want to analyze if the true mean for the amount of time spent at part time jobs is higher than 120, the system of hypothesis would be:  

Null hypothesis:\mu \leq 120  

Alternative hypothesis:\mu > 120  

Since we don't know the population deviation the statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115    

Now we can calculate the degrees of freedom

df=n-1=15-1=14  

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got t_{critc}= 1.761

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

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A sample is random sample is valid
wlad13 [49]
If the sample is random, it is not valid.
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