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3 years ago

Use the discriminant to determine the number of real solutions of the equation. x^2 = -6x - 10

Mathematics

2 answers:

HACTEHA [7]3 years ago

8 0

Answer:

No real solutions.

Step-by-step explanation:

x² + 6x + 10 = 0

a = 1

b = 6

c = 10

D = b² - 4ac = 6² - 4*1*10 = 36 - 40 = - 4

- 4 < 0,

so this equation has no real solutions.

juin [17]3 years ago

3 0

The discriminators is b^2-4ac

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Answer:

$20.9-2.01\frac{7.65}{\sqrt{50}}=18.73$

$20.9+2.01\frac{7.65}{\sqrt{50}}=23.07$

The 95% confidence interval would be given by (18.73;23.07)

Step-by-step explanation:

Assuming these data

20 40 22 22 21 21 20 10 20 20

20 13 18 50 20 18 15 8 22 25

22 10 20 22 22 21 15 23 30 12

9 20 40 22 29 19 15 20 20 20

20 15 19 21 14 22 21 35 20 22

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".

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

First we need to find the sample mean with the following formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{50}=20.9$

And in order to find the sample standard deviation we can use the following formula:

$s= \sqrt{\frac{\sum_{i=1}^n (x_i -\bar x)^2}{n-1}}=7.65$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=50-1=49$

Since the Confidence is 0.95or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that $t_{\alpha/2}=2.01$