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3 years ago

Need the answers plz

Mathematics

1 answer:

Shkiper50 [21]3 years ago

4 0

1)
         x^2 + 4x - 5
y =   --------------------
              3x^2 - 12

Vertcial asym => find the x-values for which the equation is not defined and check whether the limit goes to + or - infinite

=> 3x^2 - 12 = 0 => 3x^2 = 12

=> x^2 = 12 / 3 = 4

=> x = +/-2

Limit of y when x -> 2(+) = ( 2^2 + 4(2) - 5) / 0 = - 1 / 0(+) = ∞

Limit of y when x -> 2(-) = - 1 / (0(-) = - ∞

Limit of y when x -> - 2(+) = +∞

Limit of y when x -> - 2(-) = -

=> x = 2 and x = - 2 are a vertical asymptotes

Horizontal asymptote => find whether y tends to a constant value when x -> infinite of negative infinite

Limi of y when x -> - ∞ = 1/3

Lim of y when x -> +∞ = 1/3

=> Horizontal asymptote y = 1/3

x - intercept => y = 0

=>
          x^2 + 4x - 5
0 =   ------------------ => x ^2 + 4x - 5 = 0
              3x^2 - 12

Factor x^2 + 4x - 5 => (x + 5) (x - 1) = 0 => x = - 5 and x = 1

=> x-intercepts x = - 5 and x = 1

Domain: all the real values except x = 2 and x = - 2

2) y = - 2 / (x - 4) - 1

using the same criteria you get:

Vertical asymptote: x = 4

Horizontal asymptote: y = - 1

Domain:all the real values except x = 4

Range: all the real values except y = - 1

3)

x/ (x + 2) + 7 / (x - 5) = 14 / (x^2 - 3x - 10)

factor x^2 - 3x - 10 => (x - 5)(x + 2)

Multiply both sides by (x - 5) (x + 2)

=> x(x - 5) + 7( x + 2) = 14

=> x^2 - 5x + 7x + 14 = 14

=> x^2 + 2x = 0

=> x(x + 2) = 0 => x = 0 and x = - 2 but the function is not defined for x = - 2 so it is not a solution => x = 0

Answer: x = 0

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Step-by-step explanation:

1) Data given and notation

$\bar X=19.5$ represent the battery life sample mean

$s=1.9$ represent the sample standard deviation

$n=33$ sample size

$\mu_o =20$ represent the value that we want to test

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

t would represent the statistic (variable of interest)

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

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean battery life is less than 20 :

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Alternative hypothesis: $\mu < 20$

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:

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3) Calculate the statistic

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Since is a one-side lower test the p value would be:

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5) Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the average battery life it's not significantly different less than 20 hours at 5% of signficance. If we analyze the options given we have:

(A) Reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. FALSE, we FAIL to reject the null hypothesis.

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. TRUE, we fail to reject the null hypothesis that the mean would be 20 or higher .

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours. TRUE, we FAIL to reject the null hypothesis that the mean is greater or equal to 20 hours, so we reject the alternative hypothesis that the mean is less than 20 hours.

(D) There is enough evidence at the α=0.05 level of significance to support the claim that the true population mean battery life of the smartphone is not equal to 20 hours. FALSE the claim is not that the mean is different from 20. The real claim is: "Javier believes the mean battery life is less than 20 hours".

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