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1 year ago

This is confusing someone help please

Mathematics

1 answer:

Marat540 [252]1 year ago

4 0

Answer:

Below in BOLD.

Step-by-step explanation:

5x^2 - 6x - 2 = 0

Quadratic formula for ax^2 + bx + c = 0 is

x = [-b ± √(b^2 - 4ac)]/ 2a

Here we have a = 5, b = -6 and c = -2

so x = [-(-6) ± √((-6)^2 - 4*5*-2)]/ 2*5

= [ 6 ± √(36 + 40)] / 10

= 0.6 ± √76 / 10

= 0.6 ± 0.8718

= 1.4718, -0.2718

= 1.5, -0.3 to one decimal place.

x^2 + 3x = 40

x^2 + 3x - 40 = 0

We can factor this one:

(x - 5)(x + 8) = 0

x - 5 = 0 giving x = 5 and

x + 8 = 0 giving x = -8.

Answer x = -8, 5.

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Critical values: $t_{\alpha/2}=-2.201$ $t_{1-\alpha/2}=2.201$

95% confidence interval would be given by (3.646;4.472)

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1) 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 data is:

4.65 3.89 2.73 4.35 3.80 4.86 4.33 4.37 4.76 4.05 3.05 3.87

2) Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

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

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$

=AVERAGE(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

On this case the average is $\bar X= 4.059$

=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

The sample standard deviation obtained was s=0.6503

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =12 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by: