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zysi [14]
3 years ago
6

Given that the quadratic equation has equal roots (k^2-1)x^2-2kx-3k-1=0

Mathematics
1 answer:
Alik [6]3 years ago
7 0

Answer:

(See explanation for further details)

Step-by-step explanation:

The real expression is:

(k^{2}-1)\cdot x{{2} - 2\cdot k \cdot x - 3\cdot k + 1 = 0

The general equation for the second-order polynomial is:

x = \frac{2\cdot k \pm \sqrt{4\cdot k^{2}-4\cdot (k^{2}-1)\cdot (-3\cdot k + 1)}}{k^{2}-1}

This condition must be observed for the case of a quadratic equation with equal roots:

4\cdot k^{2} - 4\cdot (k^{2}-1)\cdot (-3\cdot k + 1) = 0

k^{2} + (k^{2}-1)\cdot (3\cdot k + 1) = 0

k^{2} + 3\cdot k^{3} - 3\cdot k - k^{2}-1 = 0

3\cdot k^{3} - 3\cdot k - 1 = 0

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Various studies indicate that approximately 11% of the world's population is left handed. You think this number is actually high
Irina-Kira [14]

Answer:

a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )

d. z= 1.3322

Step-by-step explanation:

We formulate our hypothesis as

a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )

According to the given conditions

p`= 31/225= 0.1378

np`= 225 > 5

n q` = n (1-p`) = 225 ( 1- 31/225)= 193.995> 5

p = 0.4 x= 31 and n 225

c. Using the test statistic

z=  p`- p / √pq/n

d. Putting the values

z= 0.1378- 0.11/ √0.11*0.89/225

z= 0.1378- 0.11/ √0.0979/225

z= 0.1378- 0.11/ 0.02085

z= 1.3322

at 5% significance level the z- value is ± 1.645 for one tailed test

The calculated value falls in the critical region so we reject our null hypothesis H0 : p ≤ 0.11 and accept  Ha : p >0.11 and  conclude that the data indicates that the 11% of the world's population is left-handed.

The rejection region is attached.

The P- value is calculated by finding the corresponding value of the probability of z from the z - table and subtracting it from 1.

which appears to be 0.95 and subtracting from 1 gives 0.04998

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How are adding, subtracting, dividing, and multiplying rational expressions are similar and different?
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3 years ago
Don't understand this at all
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1b) Your answer is correct.

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