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2 years ago

Find all values of k for which the equation 3x^2−4x+k=0 has no solutions.

Mathematics

2 answers:

dusya [7]2 years ago

5 0

The given equation has no solution when K is any real number and k>12

We have given that

3x^2−4x+k=0

△=b^2−4ac=k^2−4(3)(12)=k^2−144.

<h3>What is the condition for a solution?</h3>

If Δ=0, it has 1 real solution,

Δ<0 it has no real solution,

Δ>0 it has 2 real solutions.

We get,

Δ=k^2−144 here Δ is not zero.

It is either >0 or <0

Δ<0 it has no real solution,

Therefore the given equation has no solution when K is any real number.

To learn more about the solution visit:

brainly.com/question/1397278

Leona [35]2 years ago

5 0

Answer:

k>1 1/3

Step-by-step explanation:

i have rsm too bro

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Ratling [72]

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y = mx + b
y = x + 6

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2 years ago

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Answer:

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Step-by-step explanation:

Given

See attachment for graph

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To do this, we divide the sum of the products of grade and number of students by the total number of students;

i.e.

$\bar x = \frac{\sum fx}{\sum f}$

So, we have:

$\bar x = \frac{50 * 1 + 57 * 2 + 60 * 4 + 65 * 3 +72 *3 + 75 * 12 + 77 * 10 + 81 * 6 + 83 * 6 + 88 * 9 + 90 * 12 + 92 * 12 +95 * 2 + 99 * 4 + 100 * 5}{1 + 2 + 4 + 3 +3 + 12 + 10 + 6 + 6 + 9 + 12 + 12 +2 + 4 + 5}$

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Next, calculate the variance using the following formula:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f}$

i.e subtract the mean from each dataset; take the squares; add up the squares; then divide the sum by the number of dataset

So, we have:

$\sigma^2 = \frac{1 * (50 - 82.76)^2 +...................+ 5 * (100 - 82.76)^2}{91}$

$\sigma^2 = \frac{11580.6816}{91}$

$\sigma^2 = 127.26$

Lastly, take the square root of the variance to get the standard deviation

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{127.26}$

$\sigma = 11.28$ --- approximated

<em>Hence, the standard deviation is approximately 11.28</em>

Considering the calculated mean (i.e. 82.76), the standard deviation (i.e. 11.28) is small and this means that the grade of the students are close to the average grade.

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