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Leviafan [203]
3 years ago
13

Hey can anyone help me out in dis!!!!

Mathematics
2 answers:
Novay_Z [31]3 years ago
5 0
5) 2(6x - 5) < 14
12x - 10 < 14
12x < 24
x < 2
6) 5(x-2)/10 > 6
10*5(x-2)/10 > 6*10
50(x-2) > 60
50x - 100 > 60
50x > 160
x > 3.2
7) 15>= 3(7x-9)
15 >= 21x - 27
42 >= 21x
2 >= x
Or x <= 2
8) x+14 < x/2
2(x+14) < x
2x + 28 < x
x < -28
Semenov [28]3 years ago
5 0

Answer:

Step-by-step explanation:

5.

2\left(6x-5\right)

6.

\frac{5\left(x-2\right)}{10}>6\\\mathrm{Multiply\:both\:sides\:by\:}10\\\frac{10\cdot \:5\left(x-2\right)}{10}>6\cdot \:10\\Simplify\\5\left(x-2\right)>60\\\mathrm{Divide\:both\:sides\:by\:}5\\\frac{5\left(x-2\right)}{5}>\frac{60}{5}\\Simplify\\x-2>12\\\mathrm{Add\:}2\mathrm{\:to\:both\:sides}\\x-2+2>12+2\\x>14

8.

x+14

7.

15\ge \:3\left(7x-9\right)\\Switch\:sides\\3\left(7x-9\right)\le \:15\\\mathrm{Divide\:both\:sides\:by\:}3\\\frac{3\left(7x-9\right)}{3}\le \frac{15}{3}\\Simplify\\7x-9\le \:5\\\mathrm{Add\:}9\mathrm{\:to\:both\:sides}\\7x-9+9\le \:5+9\\Simplify\\7x\le \:14\\\mathrm{Divide\:both\:sides\:by\:}7\\\frac{7x}{7}\le \frac{14}{7}\\\mathrm{Simplify}\\x\le \:2

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Answer:

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Answer:

z=\frac{704-732}{\frac{65}{\sqrt{28}}}=-2.279      

p_v =P(z  

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we can reject the null hypothesis, and the the true mean is significantly lower than 732 hours so we have enough evidence to reject the claim

Step-by-step explanation:

Data given and notation      

\bar X=704 represent the sample mean

\sigma=65 represent the standard deviation for the population      

n=28 sample size      

\mu_o =732 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)      

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the true mean is at least 732 or no, the system of hypothesis would be:      

Null hypothesis:\mu \geq 732      

Alternative hypothesis:\mu < 732      

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:      

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)      

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic      

We can replace in formula (1) the info given like this:      

z=\frac{704-732}{\frac{65}{\sqrt{28}}}=-2.279      

Calculate the P-value      

Since is a one-side lower test the p value would be:      

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Conclusion      

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we can reject the null hypothesis, and the the true mean is significantly lower than 732 hours so we have enough evidence to reject the claim

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Answer:

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Step-by-step explanation:

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Finally, we can see that the value of k is 20,000 per sq. km.

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