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Mamont248 [21]
2 years ago
13

Consider the equation y = a - bx. If y = 6 when x = 1 and y = 0 when x = 2, find . = х

Mathematics
1 answer:
Trava [24]2 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

(a)

y = \frac{a}{x} - bx

substitute y = 6 , x = 1 into the equation

6 = a - b → (1)

substitute y = 0 , x = 2 into the equation

0 = \frac{a}{2} - 2b ( multiply through by 2 to clear the fraction )

0 = a - 4b → (2)

multiply (1) by - 4

- 24 = - 4a + 4b → (3)

add (2) and (3) term by term to eliminate b

- 24 = - 3a + 0

- 24 = - 3a ( divide both sides by - 3 )

8 = a

substitute a = 8 into (1) and solve for b

6 = 8 - b ( subtract 8 from both sides )

- 2 = - b ( multiply both sides by - 1 )

2 = b

Then a = 8 and b = 2

(b)

when x = 4

y = \frac{8}{4} - 2(2) = 2 - 4 = - 2

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Answer:

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We need to find the degrees of freedom given by:

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Since is a right tailed test the p value would be:  

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Step-by-step explanation:

Data given and notation  

\bar X=54.98 represent the sample mean

s=8.43 represent the sample deviation

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\mu_o =52 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 52, the system of hypothesis would be:  

Null hypothesis:\mu \leq 52  

Alternative hypothesis:\mu > 52  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-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".  

Calculate the statistic  

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

t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025  

P-value  

We need to find the degrees of freedom given by:

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Since is a right tailed test the p value would be:  

p_v =P(t_{25}>1.8025)=0.0418  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.  

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