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iris [78.8K]
3 years ago
7

Solve the inequality 4/3 < -8/3y

Mathematics
1 answer:
natima [27]3 years ago
6 0

Answer:

y < -1/2

Step-by-step explanation:

Step  1  :

           8

Simplify   —

           3

Equation at the end of step  1  :

 4           8

 — -  (0 -  (— • y))  < 0

 3           3

Step  2  :

           4

Simplify   —

           3

Equation at the end of step  2  :

 4    -8y

 — -  ———  < 0

 3     3

Step  3  :

Adding fractions which have a common denominator :

3.1       Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

4 - (-8y)     8y + 4

—————————  =  ——————

    3           3  

Step  4  :

Pulling out like terms :

4.1     Pull out like factors :

  8y + 4  =   4 • (2y + 1)

Equation at the end of step  4  :

 4 • (2y + 1)

 ————————————  < 0

      3      

Step  5  :

5.1    Multiply both sides by  3

5.2    Divide both sides by  4

5.3    Divide both sides by  2  

     y+(1/2)  < 0

Solve Basic Inequality :

5.4      Subtract  1/2  from both sides

           y < -1/2

Inequality Plot :

5.5      Inequality plot for

        2.667 X  + 1.333  <  0

 

One solution was found :

                  y < -1/2

Processing ends successfully

plz mark me as brainliest if this helped :)

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H=hour
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3 years ago
A TV station claims that 38% of the 6:00 - 7:00 pm viewing audience watches its evening news program. A consumer group believes
Elena L [17]

Answer:

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

p_v =P(Z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

Step-by-step explanation:

1) Data given and notation

n=830 represent the random sample taken

X=282 represent the people that regularly watch the TV station’s news program

\hat p=\frac{282}{830}=0.340 estimated proportion of people that regularly watch the TV station’s news program

p_o=0.38 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.38:  

Null hypothesis:p\geq 0.38  

Alternative hypothesis:p < 0.38  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

3 0
3 years ago
Air-USA has a policy of booking as many as 24 persons on an airplane that can seat only 22. (Past studies have revealed that onl
Alex17521 [72]

Answer:

B. no, it is not low enough

A. no, it is not low enough

Step-by-step explanation:

Given that Air-USA has a policy of booking as many as 24 persons on an airplane that can seat only 22.

Prob for  a random person booked arrive for flight = 0.86

No of persons who books and arrive for flight, X is binomial, since there are two outcomes and each person is independent of the other

The probability that if Air-USA books 24 persons, not enough seats will be available

= P(X=23)+P(x=24)

= 0.1315

B. no, it is not low enough

-------------------------------

The prob we got is >10% also

A. no, it is not low enough

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ch4aika [34]

Answer:

The population of Bear in 2050 is 4750000

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dN/dT = rmax * N

Where dN/dT = change in population

rmax is the maximum rate of change

N = Base population

B) Here the per capita rate of increase (r) will always be a positive value irrespective of the and hence we will assume this population to be growing exponentially.

C) dN/dT = rmax * N

D) dN / 5 = 2.5 * 380,000

dN = 5*2.5 * 380000

= 4750000

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1/2 divided by 3 is 1/6.
So your answer is 1/6 pounds each person.
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3 years ago
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