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3 years ago

How do I figure out X in this equation URGENT!!!

Mathematics

2 answers:

Mekhanik [1.2K]3 years ago

7 0

2*(X-2+3x+5)=94 cm
4x+3 = 47
4x=44
x=11 cm

Mama L [17]3 years ago

6 0

Step-by-step explanation:

p = 2(L+W)

98 < 2(x-2cm + 3x+5cm)

98 < 2x - 4 + 6x + 10

98 < 2x + 6x -4 +10

98 < 8x + 6........the inequality

98 - 6 < 8x

92 < 8x

11.5 < x.......the value of x

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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 - α)% 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:

n = 20

s = 3.9

Confidence level = 95%

⇒ α = 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).

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5 0

3 years ago

Read 2 more answers

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spayn [35]

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The formula for a cylinder is 3.14*r^2*h but if you don't round to the nearest tenth the the answer is 197.82034.

Step-by-step explanation:Step one mutiply 3^2=9, Step two mutiply 3.14*

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3 years ago