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tankabanditka [31]
3 years ago
6

Y’all I need help on #9

Mathematics
1 answer:
Artist 52 [7]3 years ago
7 0

Answer:

8° = y

12° = x

________________

11° = z

7° = y

23 = x

Step-by-step explanation:

<u>What</u><u> </u><u>you</u><u> </u><u>need</u><u> </u><u>to know</u>

- m∠K ≅ m∠W; KB = WM

- m∠G ≅ m∠T; KG = WT

- m∠B ≅ m∠M; GB = TM

45° = [4x - 3]°

+ 3° + 3°

____________

\frac{48°}{4°} = \frac{[4x]°}{4°} \\ \\ 12° = x

Then use the Triangular Interior Angles Theorem to find the m∠Mthen set that equal to the m∠B:

180° = 41° + 45° + m∠M

180° = 86° + m∠M

- 86° - 86°

______________

94° = m∠M

94° = [11y + 6]°

- 6° - 6°

__________

\frac{88°}{11°} = {[11y]°}{11°} \\ \\ 8° = y

_______________________________________________

<u>What</u><u> </u><u>you</u><u> </u><u>need</u><u> </u><u>to know</u>

- m∠H ≅ m∠S; HC = SP

- m∠F ≅ m∠R; CS = RP

- m∠C ≅ m∠P; HF = SR

90° = [13y - 1]°

+ 1° + 1°

______________

\frac{91°}{13°} = \frac{[13y]°}{13°} \\ \\ 7° = y \\ \\ 90° = m∠R

Then use the Triangular Interior Angles Theorem to find the m∠Sthen set that equal to the m∠H:

180° = 28° + 90° + m∠S

180° = 118° + m∠S

- 118° - 118°

______________

62° = m∠S

62° = [6z - 4]°

+ 4° + 4°

____________

\frac{66°}{6°} = \frac{[6z]°}{6°} \\ \\ 11° = z

I am joyous to assist you anytime.

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(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

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(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

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Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

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Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

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Step-by-step explanation:

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