All categories

2 years ago

Can someone plz help me with this one problem

Mathematics

1 answer:

galina1969 [7]2 years ago

5 0

The awswer should be 18? If it’s adding to find.

You might be interested in

15 DIVIDED 6 2/3 A. 2 3/4 B.100 C.100 1/4 D.2 1/4 HELP!!!!!!

Andrew [12]

D, 2 and 1/4 or 2.25

4 0

3 years ago

Please help me with this i’m stuck

bagirrra123 [75]

Answer:

A ) Substitute the side lengths of a and b in the equation

Step-by-step explanation:

The hypotenuse which is opposite the right angle it always the length of C in the cosine rule.

5 0

2 years ago

To which graph does the point (8, −2) belong?

Leni [432]

D because it is 2<17

4 0

3 years ago

Read 2 more answers

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

2 years ago

How many radians is 105°?

Katen [24]

Answer:

The answer is 1.833 radians

7 0

3 years ago

Read 2 more answers