How many solutions does the system of equations below have? x+y=4 and x-y=6

Over [174]

Looking at this we can group like terms. What are like terms? They are terms that have the same variable that we can add or subtract. In this case the like terms are 6m and m because they both have the same variable and we could add them together to get 7m.

6 0

3 years ago

50 points Type the correct answer I’m the box

Nonamiya [84]

I only see a black image, can you resend the image? or type it out here

6 0

3 years ago

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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

Round 299.2131 to the nearest thousands

Digiron [165]

Answer:

Step-by-step explanation:

First find the place that is the thousandth place. That would be the 3 then you will look beside the 3 to see if their is a number bigger than five. If there is then you would add whatever number to make the number 10. In this case their is not so you would instead look at the three and replace all the numbers to the right of three ( don’t change three though) and make them zeros. So that will leave you with 299.2130 rounded to the nearest thousandth.

5 0

3 years ago

Question 2 (10 points)

lara [203]

Answer:

140-111=29

Step-by-step explanation:

5 0

3 years ago

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