The equation x^3-5x=59 has a solution between 4and 5 . Use trail and improvement method. Give your answer to 1 decimal place.

Mathematics

1 answer:

3241004551 [841]3 years ago

5 0

Answer:

4.3 to 1 decimal place.

Step-by-step explanation:

x^3 - 5x - 59 = 0

Try x = 4.1:

f(4.1) = (4.1)^3 - 5(4.1) - 59 = -10.579

Try: 4.2:

f(4.2) = 4.2^3 - 5(4.2) - 59 = -5.912, Try x = 4.3:

f(4.3) = -0.993 (getting closer to 0!!)

f(4.4) = 4.184

NOTICE - there is a change of sign in the result so the root lies between 4.3 and 4.4.

Try x = 4.35:

f(4.35) = 1.563 so it looks like the root is closer to 4.3 than 4.4 because 1.563 is closer to zero than 4.184.

Just to double check let's calculate x = 4.32:

f(4.32) = 0.0216 - which is very close to zero.

So the answer is 4.3 to 1 decimal place.

(I've checked it out on my calculator - the root is 4.31958).

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My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the

AVprozaik [17]

Answer:

a) $n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

b) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by: