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2 years ago

If a = 5+2√6 then find a²-1÷a²

Mathematics

1 answer:

Umnica [9.8K]2 years ago

5 0

Answer:

a²+1/a² = 7

Step-by-step explanation:

Given, a = (3+√5)/2

We have to find the value of a² + 1/a².

1/a = 1/((3+√5)/2) = 2/(3+√5)

By taking conjugate,

2/(3+√5) = 2/(3+√5) × (3-√5)/(3-√5)

= 2(3-√5) / (3+√5)(3-√5)

By using algebraic identity,

(a - b)(a + b) = a² - b²

(3+√5)(3-√5) = (3)² - (√5)²

= 9 - 5

= 4

2(3-√5) / (3+√5)(3-√5) = 2(3-√5) / 4

So, 1/a = (3-√5)/2

Now, a² = [(3+√5)/2]²

= (3+√5)²/4

By using algebraic identity,

(a + b)² = a² + 2ab + b²

(3+√5)² = (3)² + 2(3)(√5) + (√5)²

= 9 + 6√5 + 5

= 14 + 6√5

a² = (14+6√5)/4

= 2(7+3√5)/4

a² = (7+3√5)/2

Now, 1/a² = [(3-√5)/2]²

= (3-√5)²/4

By using algebraic identity,

(a - b)² = a² - 2ab + b²

(3-√5)² = (3)² - 2(3)(√5) + (√5)²

= 9 - 6√5 + 5

= 14 - 6√5

1/a² = (14-6√5)/4

1/a² = (7-3√5)/2

a²+1/a² = (7+3√5)/2 + (7-3√5)/2

= 7/2 + 3√5/2 + 7/2 - 3√5/2

= 7/2 + 7/2

= 7

Therefore, a²+1/a² = 7

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(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).

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Step-by-step explanation:

Let X = temperature of a certain solution.

The estimated mean temperature is, $\bar x=37^{o}C$ .

The estimated standard deviation is, $s=0.1^{o}C$ .

(a)

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Since the sample size is large use a z-confidence interval.

The critical value of z for 95% confidence interval is:

$z_{0.025}=1.96$

Compute the confidence interval as follows:

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Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).

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The confidence interval is, 37 ± 0.1°C.

Comparing the confidence interval with the general form:

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The critical value is,

CV = 1

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The percentage of z-distribution between -1 and 1 is, 68%.

Thus, the confidence level is 68%.

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The confidence interval for population mean can be constructed using either the z-interval or t-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a t-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

(d)

For n = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - α)% t-confidence interval is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

The critical value of t is:

$t_{\alpha/2, (n-1)}=t_{0.025, 9}=2.262$

*Use a t-table for the critical value.

The 95% confidence interval is:

$CI=37\pm 2.262\times \frac{0.1}{\sqrt{10}}\\=37\pm 0.072\\=(36.928, 37.072)\\\approx (36.93^{o}C, 37.07^{o}C)$

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