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posledela
3 years ago
10

Solve the equation below on the interval [0, 2pi). Select ALL that apply.

Mathematics
1 answer:
tia_tia [17]3 years ago
7 0
D is the answer I hope it help
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More 5th grade work c’mon
fgiga [73]

Answer: honestly for the first one, i think the answer would be 2.

they both have 5 at the end so you know its divisible :)

Step-by-step explanation:

6 0
3 years ago
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
Find the volume of a cube where the length of each side is 6 centimeters.
Marrrta [24]

hope it will help u..........

6 0
3 years ago
Roy is twice as old as Joan, and in 3 years the sum of their ages will be 21 years. Find their present ages.
Lisa [10]

Answer:

Roy is 10 years old at present and Joan is 5 years old at present

Step-by-step explanation:

Let

x----> Roy's age

y----> Joan's age

we know that

x=2y ----> equation A

(x+3)+(y+3)=21 ----> equation B

substitute equation A in equation B

(2y+3)+(y+3)=21

solve for y

3y+6=21

3y=21-6

3y=15

y=5 years

Find the value of x

x=2y  ----> x=2(5)=10 years

therefore

Roy is 10 years old at present

Joan is 5 years old at present

8 0
3 years ago
X + 2y = -4 in graph
MariettaO [177]

Equation Answer:

y = -½x - 2

Graph:

3 0
3 years ago
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