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vagabundo [1.1K]
3 years ago
12

Which is a solution of x2 – x – StartFraction 3 Over 4 EndFraction = 0?

Mathematics
2 answers:
V125BC [204]3 years ago
8 0

Answer:

3/2

Step-by-step explanation:

Anna [14]3 years ago
4 0

Answer:

x = - \frac{1}{2}, x = \frac{3}{2}

Step-by-step explanation:

Given

x² - x - \frac{3}{4} = 0 ( add \frac{3}{4} to both sides )

x² - x = \frac{3}{4}

Solve using the method of completing the square

add ( half the coefficient of the x- term )² to both sides

x² + 2( - \frac{1}{2} )x + \frac{1}{4} = \frac{3}{4} + \frac{1}{4}

(x - \frac{1}{2} )² = 1 ( take the square root of both sides )

x - \frac{1}{2} = ± 1 ( add \frac{1}{2} to both sides )

x = \frac{1}{2} ± 1

Thus

x = \frac{1}{2} - 1 = - \frac{1}{2}

x = \frac{1}{2} + 1 = \frac{3}{2}

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Answer:

a) p_v =P(Z          

b) Since p_v we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) P(\bar X >90)=1-P(Z

1-P(Z

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

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Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

We can calculate the sample mean and deviation with the following formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

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s=1.011 represent the sample standard deviation  

n=5 represent the sample selected  

\alpha significance level    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:    

Null hypothesis:\mu \geq 90    

Alternative hypothesis:\mu < 90    

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

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We can replace in formula (1) the info given like this:    

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Part a

P-value  

First we need to calculate the degrees of freedom given by:

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Then since is a left tailed test the p value would be:    

p_v =P(Z    

Part b

Since p_v we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

P(\bar X

The best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

If we apply this formula to our probability we got this:

P(\bar X >90)=1-P(Z

1-P(Z

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2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}

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