Solve for the missing values in the table y=3x-2

Estimate the measure of the clockwise angle formed by the minute hand and the second hand, then classify the angle. A. 54°; acut

Harlamova29_29 [7]

I would go with b or a or d but not c

8 0

3 years ago

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Which of these sets of angle measures could be three angles in a triangle

erastovalidia [21]

It’s going to be B. 50+60+70=180 which is how many degrees are in a triangle.

8 0

3 years ago

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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).

4 0

3 years ago

HELP PLZ! I do not understand this!!

neonofarm [45]

PART 1
If I have graph f(x) then graph f(x + 1/3) would translate the graph 1/3 to the left.

For example, I have f(x) = x².Then
f(x + 1/3) = (x + 1/3)²

I draw the graph of f(x) = x² and the graph of f(x) = (x + 1/3)² on cartesian plane to know what's the difference between them.

PART 2
If I have graph f(x) then graph f(x) + 1/3 would translate the graph 1/3 upper.

For example, I have f(x) = x².Then
f(x) + 1/3 = x² + 1/3

I draw the graph of f(x) = x² and the graph of f(x) = x² + 1/3 on cartesian plane to know what's the difference between them.

SUMMARY
f(x+1/3) ⇒⇒ f(x) is translated 1/3 units left.
f(x) + 1/3 ⇒⇒ f(x) is translated 1/3 units up.