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KengaRu [80]
3 years ago
9

The angle by which AB turns clockwise about point B to coincide with BC is ? degrees. If from point B, a point E is drawn direc

tly opposite point C so that B, E, and C are on the same straight line, the angle by which AB turns counterclockwise to coincide with BE is ? degrees.
fill in the ? marks

Mathematics
2 answers:
BigorU [14]3 years ago
7 0

Answer:

Step-by-step explanation:

(A) it is given that if AB turns clockwise about point B to coincide with BC, then the angle will be=∠ABD+∠DBC=33.3°+30.6°=63.9°.

(B) If from point B, a point E is drawn directly opposite point C so that B, E, and C are on the same straight line, then the angle by which AB turns clockwise to coincide with BE is:∠ABE+∠ABD+∠DBC=180° (using the straight line property)

⇒∠ABE+33.3°+30.6°=180°

⇒∠ABE+63.9°=180°

⇒∠ABE=116.1°

Oksana_A [137]3 years ago
3 0
A) AB turn clockwise to coincide with BC

= ABD + DBC

= 33.3° + 30.6° 

= 63.9°


b) If E is drawn directly opposite C.

EBC is a straight line, so sum of angles =180°

ABE + ABD + DBC = 180°

ABE + 33.3° + 30.6° = 180°

ABE + 63.9° = 180°

ABE = 180 - 63.1

ABE = 116.9° 

Hope this explains it.
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A university found that of its students withdraw without completing the introductory statistics course. Assume that students reg
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Answer:

A university found that 30% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course.

a. Compute the probability that 2 or fewer will withdraw (to 4 decimals).

= 0.0355

b. Compute the probability that exactly 4 will withdraw (to 4 decimals).

= 0.1304

c. Compute the probability that more than 3 will withdraw (to 4 decimals).

= 0.8929

d. Compute the expected number of withdrawals.

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Step-by-step explanation:

This is a binomial problem and the formula for binomial is:

P(X = x) = nCx p^{x} q^{n - x}

a) Compute the probability that 2 or fewer will withdraw

First we need to determine, given 2 students from the 20. Which is the probability of those 2 to withdraw and all others to complete the course. This is given by:

P(X = x) = nCx p^{x} q^{n - x}\\P(X = 2) = 20C2(0.3)^2(0.7)^{18}\\P(X = 2) =190 * 0.09 * 0.001628413597\\P(X = 2) = 0.027845872524

P(X = x) = nCx p^{x} q^{n - x}\\P(X = 1) = 20C1(0.3)^1(0.7)^{19}\\P(X = 1) =20 * 0.3 * 0.001139889518\\P(X = 1) = 0.006839337111

P(X = x) = nCx p^{x} q^{n - x}\\P(X = 0) = 20C0(0.3)^0(0.7)^{20}\\P(X = 0) =1 * 1 * 0.000797922662\\P(X = 0) = 0.000797922662

Finally, the probability that 2 or fewer students will withdraw is

P(X = 2) + P(X = 1) + P(X = 0) \\= 0.027845872524 + 0.006839337111 + 0.000797922662\\= 0.035483132297\\= 0.0355

b) Compute the probability that exactly 4 will withdraw.

P(X = x) = nCx p^{x} q^{n - x}\\P(X = 4) = 20C4(0.3)^4(0.7)^{16}\\P(X = 4) = 4845 * 0.0081 * 0.003323293056\\P(X = 4) = 0.130420974373\\P(X = 4) = 0.1304

c) Compute the probability that more than 3 will withdraw

First we will compute the probability that exactly 3 students withdraw, which is given by

P(X = x) = nCx p^{x} q^{n - x}\\P(X = 3) = 20C3(0.3)^3(0.7)^{17}\\P(X = 3) = 1140 * 0.027 * 0.002326305139\\P(X = 3) = 0.071603672205\\P(X = 3) = 0.0716

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d) Compute the expected number of withdrawals.

E(X) = 3/10 * 20 = 6

Expected number of withdrawals is the 30% of 20 which is 6.

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