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puteri [66]
3 years ago
10

Y=.5x+4 Y= -.25x^2+5 Using the quadratic formula, what are the solutions?

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Hi there!

\large\boxed{x = -1 \pm \sqrt{5}}

We can solve by setting both equations equal to each other:

.5x + 4 = -.25x² + 5

Move everything to one side:

0 = -.25x² - .5x + 1

Multiply both sides by 4 to cancel out the decimals:

0 = -x² - 2x + 4

Factor out negative from both sides:

0 = x² + 2x - 4

Use the quadratic formula:

\frac{-b \pm \sqrt{b^2-4ac}}{2a}

Plug in the corresponding terms:

x = \frac{-2 \pm \sqrt{4-4(-4)(1)}}{2}

Simplify:

x = \frac{-2 \pm \sqrt{4 + 16}}{2}\\\\x = \frac{-2 \pm \sqrt{20}}{2}

Simplify the radical to reduce:

x = \frac{-2 \pm 2\sqrt{5}}{2}

Divide all terms by 2:

x = -1 \pm \sqrt{5}

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Answer:

p_v =P(z>3.390)=0.000349  

If we compare the p value and the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of returns at the Houston store is significantly higher than 0.06 or 6%.

Step-by-step explanation:

1) Data given and notation n  

n=80 represent the random sample taken

X=12 represent number of items returned

\hat p=\frac{12}{80}=0.15 estimated proportion of items returned

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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 proportion of returns at the Houston store was more than the national expectation (0.06):  

Null hypothesis:p\leq 0.06  

Alternative hypothesis:p >0.06  

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{0.15 -0.06}{\sqrt{\frac{0.06(1-0.06)}{80}}}=3.390  

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=0.1. 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>3.390)=0.000349  

If we compare the p value and the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of returns at the Houston store is significantly higher than 0.06 or 6%.

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