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3 years ago

Solve the equation for x. If necessary round to the nearest tenth. x² = 17 X= +

Mathematics

1 answer:

Ilya [14]3 years ago

7 0

Answer:

x = 4.1 & x = −4.1

Step-by-step explanation:

$x^2 = 17\\$

x = ±√17

x = √17, −√17

x = 4.1 & x = −4.1

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Which of the following triangles are similar?

lana66690 [7]

A is the answer to your question

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3 years ago

Simplify the expression and write your answer in scientific notation: (1.9 X 10^5)^3

lord [1]

Answer:

${(1.9 \times {10}^{5}) }^{3}$

open the bracket:

$= {1.9}^{3} \times {10}^{(5 \times 3)} \\ { \boxed{= 6.859 \times {10}^{15} }}$

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3 years ago

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Misha Larkins [42]

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3 years ago

Read 2 more answers

In a multiple regression equation, two independent variables are considered, and the sample size is 30. The regression coefficie

MAXImum [283]

Answer:

Both the variables are important for the regression analysis and cannot be deleted.

Step-by-step explanation:

(1)

The hypothesis for for the testing of coefficient β₁ are:

H₀: β₁ = 0 vs. Hₐ: β₁ ≠ 0

The test statistic is:

$t=\frac{b_{1}}{S_{b_{1}}} =\frac{2.815}{0.75}= 3.753$

It is provided that H₀ is rejected if t > 2.042.

The test statistic value, t = 3.753 > 2.042.

Thus, the null hypothesis is rejected.

Conclusion:

There is a significant relationship between the regression variable and the dependent variable.

(2)

The hypothesis for for the testing of coefficient β₂ are:

H₀: β₂ = 0 vs. Hₐ: β₂ ≠ 0

The test statistic is:

$t=\frac{b_{2}}{S_{b_{2}}} =\frac{-1.249}{0.41}= -3.046$

It is provided that H₀ is rejected if t < -2.042.

The test statistic value, t = -3.046 < -2.042.

Thus, the null hypothesis is rejected.

Conclusion:

There is a significant relationship between the regression variable and the dependent variable.

Thus, both the variables are important for the regression analysis and cannot be deleted.

7 0

3 years ago

Himpunan penyelesaian pertidaksamaan nilai mutlak |2x-4| > |3x+2|

Stells [14]

Answer:

The solution in interval notation is:

$(-6,\frac{2}{5})$ .

The solution in inequality notation is:

$-6$ .

Step-by-step explanation:

I think you are asking how to solve this for $x$ .

Keep in mind $|x|=\sqrt{x^2}$ .

$|2x-4|>|3x+2|$

$\sqrt{(2x-4)^2}>\sqrt{(3x+2)^2}$

If $\sqrt{u}>\sqrt{v}$ then $u>v$ .

$(2x-4)^2>(3x+2)^2$

Subtract $(2x-4)^2$ on both sides:

$0>(3x+2)^2-(2x-4)^2$

Factor the difference of squares $a^2-b^2=(a-b)(a+b)$ :

$0>((3x+2)-(2x-4))((3x+2)+(2x-4))$

Simplify inside the factors:

$0>(x+6)(5x-2)$

$(x+6)(5x-2)$

The left hand side is a parabola that faces up. I know this because the degree is 2.

The zeros of the the parabola are at x=-6 and x=2/5.

We can solve x+6=0 and 5x-2=0 to reach that conclusion.

x+6=0

Subtract 6 on both sides:

x=-6

5x-2=0

Add 2 on both sides:

5x=2

Divide both sides by 5:

x=2/5

Since the parabola faces us and $(x+6)(5x-2)$ then we are looking at the interval from x=-6 to x=2/5 as our solution. That part is where the parabola is below the x-axis. We are looking for where it is below since it says the where is the parabola<0.

The solution in interval notation is:

$(-6,\frac{2}{5})$ .

The solution in inequality notation is:

$-6$ .

6 0

4 years ago