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Mekhanik [1.2K]
3 years ago
14

The common point between lines y = 2x + 5 and y = ½ x + 6 is (3, 1/2). true or false

Mathematics
2 answers:
Taya2010 [7]3 years ago
8 0

Answer:

the answer is false.

Step-by-step explanation:


mixer [17]3 years ago
5 0
The common point between y = 2x + 5 and y = (1/2)x + 6 will have the same values for x and y.

Therefore, set the two y expressions equal to obtain
2x + 5 = (1/2)x + 6

Subtract (1/2)x from each side.
(3/2)x + 5 = 6

Subtract 5 from each side.
(3/2)x = 1

Multiply each side by 2/3.
x = 2/3.

From the first equation, obtain
y = 2*(2/3) + 5 = 19/3.

The common point is (2/3, 19/3). It is not equal to (3, 1/2).

Answer: False
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A newspaper poll asked respondents if they trusted "eco friendly" labels on cleaning products. Out of 1000 adults surveyed, 498
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Answer:

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of respondents that trust these labels is not significanlty less than 0.5 or 50%.  

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=498 represent the adults that trust these labels

\hat p=\frac{498}{1000}=0.498 estimated proportion of respondents that trust these labels

p_o=0.5 is the value that we want to test

\alpha represent the significance level

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 of respondents that trust these labels is at least 50%:  

Null hypothesis:p\geq 0.5  

Alternative hypothesis:p < 0.5  

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{0.498 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=-0.126  

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 is not provided but we can assume it as \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 high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of respondents that trust these labels is not significanlty less than 0.5 or 50%.  

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