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diamong [38]
3 years ago
5

Solve: 3[-x+(2x=1)]=x-1 I can't distribute so I don't know what to do.

Mathematics
1 answer:
Nat2105 [25]3 years ago
8 0

Answer:

Below

Step-by-step explanation:

● 3 ( -x +2x - 1) = x - 1

● 3 ( -x -1 ) = x - 1

● -3x -3 = x - 1

Add -x to both sides

● -3x -3 - x = x -1 -x

● -4x -3 = -1

Add 3 to both sides

● -4x - 3 + 3 = -1 + 3

● -4x = 2

Divide both sides by -4

● -4x/-4 = 2/-4

● x = -1/2

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Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in
Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population variance is given as follows:

\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}

It is provided that:

<em>n</em> = 20

<em>s</em> = 3.9

Confidence level = 95%

⇒ <em>α</em> = 0.05

Compute the critical values of Chi-square:

\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}

\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

4 0
3 years ago
What is the simplified form of 12x/900x²y4z6?
zalisa [80]

Answer:

360xy^2|xz^3|

Step-by-step explanation:

Given expression:

12x \sqrt{900x^2y^4z^6}

\textsf{Apply radical rule} \quad \sqrt{ab}=\sqrt{a}\sqrt{b}:

\implies 12x \sqrt{900}\sqrt{x^2}\sqrt{y^4}\sqrt{z^6}

Replace 900 with 30² :

\implies 12x \sqrt{30^2}\sqrt{x^2}\sqrt{y^4}\sqrt{z^6}

\textsf{Apply radical rule} \quad \sqrt{a^2}=a, \quad a \geq 0:

\implies 12x \cdot 30|x|\sqrt{y^4}\sqrt{z^6}

\implies 360x|x|\sqrt{y^4}\sqrt{z^6}

(We need to use the absolute value of √x² since the x term was originally to the power of 2, which means the value of x² is always positive since the exponent is even).

\textsf{Apply exponent rule} \quad \sqrt{a^m}=a^{\frac{m}{2}}:

\implies 360x|x|\cdot y^{\frac{4}{2}}\cdot z^{\frac{6}{2}}

Simplify:

\implies 360xy^2|xz^3|

(We need to use the absolute value of z³ since the z term was original to the power of 6, which means the value of z⁶ is always positive since the exponent is even).

7 0
2 years ago
, list the first five terms of each sequence, and identify them as arithmetic or geometric.
Natali [406]

Answer:

Step-by-step explanation:

A_{n+1}=14*A_{n}

A_{1}=8

A_{1+1}=14×8=112

A_{2}=112

A_{3}=14*A_{2}=14×112=1568

A_{4}=14*A_{3}=14×1568=21952

A_{5}=14*A_{4}=14×21952=307328

7 0
3 years ago
Y=(x-5)(x+1). What is the vertex of the parabola
dedylja [7]

Answer:

Rewrite in vertex form and use this form to find the vertex  

(h , k ) .  ( 2 , − 9 )

5 0
3 years ago
Which one do I put? Pick out of the 3 options.
SpyIntel [72]
The answer is -6 is less than 3
3 0
3 years ago
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