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Montano1993 [528]
2 years ago
14

PLEASE HELP QUICK 20 POINTS AND BEST ANSWER

Mathematics
1 answer:
Lostsunrise [7]2 years ago
3 0

Answer:

not a function. hope this helps :D

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Solve 2(1 – x) &gt; 2x<br> this is my question for algebra
Rudik [331]

To solve this, you need to isolate/get the variable "x" by itself in the inequality:

2(1 - x) > 2x     Divide 2 on both sides

\frac{2(1-x)}{2} >\frac{2x}{2}

1 - x > x     Add x on both sides to get "x" on one side of the inequality

1 - x + x > x + x

1 > 2x        Divide 2 on both sides to get "x" by itself

\frac{1}{2} >x  or  x     (x is any number less than 1/2)

[Another way you could've solved it]

2(1 - x) > 2x     Distribute 2 into (1 - x)

(2)1 + (2)(-x) > 2x

2 - 2x > 2x      Add 2x on both sides

2 - 2x + 2x > 2x + 2x

2 > 4x    Divide 4 on both sides to get "x" by itself

\frac{2}{4} >\frac{4x}{4}

\frac{1}{2} >x

6 0
2 years ago
Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n
Oksanka [162]

Answer:

z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

p_v =P(Z>2.097)=0.018  

If we compare the p value obtained and 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  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

\hat p=\frac{209}{377}=0.554 estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

p_o=0.5 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 true proportion is higher than 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statistic, 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{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

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 right tailed test the p value would be:  

p_v =P(Z>2.097)=0.018  

If we compare the p value obtained and 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  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

6 0
3 years ago
Drag the simplified value into the box to match each expression.
AveGali [126]

Answer:

2^{3} +1⋅4−3 = 9

2^{0}+10−4⋅2 = 3

3+5^{2} -6⋅4 = 4

Step-by-step explanation:

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2 years ago
How many times larger is a centigram than a milligram
Sauron [17]

Answer:

centi = a unit / 100

milli = a unit / 1000

centiX = 10 milliX

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3 years ago
Leon got a reduction of $40 on a Geometry set
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I pretty sure this one is b
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