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3 years ago

HELP ME GUYS I AM ILLITERATE find the size of unknown angle

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

5 0

Answer:

Step-by-step explanation:

ΔAOC is an isosceles triangle as OC = OA ----> RADIUS

42 + 42 + ∠AOC = 180 {Angle sum property}

84 + ∠AOC = 180

∠AOC = 180 - 84

∠AOC = 96°

y = 96/2 {Central angle theorem}

y = 48° ------------(I)

∠B + ∠D = 180 {sum of opposite angles of Cyclic Quadrilateral}

x + y = 180

x + 48 = 180 {from (I) }

x = 180 - 48

x = 132°

Vlada [557]3 years ago

3 0

Answer:

x = 132°

Step-by-step explanation:

ΔAOC is an isosceles triangle as OC = OA ----> RADIUS

42 + 42 + ∠AOC = 180 {Angle sum property}

84 + ∠AOC = 180

∠AOC = 180 - 84

∠AOC = 96°

y = 96/2 {Central angle theorem}

y = 48° ------------(I)

∠B + ∠D = 180 {sum of opposite angles of Cyclic Quadrilateral}

x + y = 180

x + 48 = 180 {from (I) }

x = 180 - 48

x = 132°

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loris [4]

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8 0

3 years ago

Read 2 more answers

A university found that of its students withdraw without completing the introductory statistics course. Assume that students reg

polet [3.4K]

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.

= 6

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

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

Then, using a) we have that the probability that 3 or fewer students withdraw is 0.0355+0.0716=0.1071. Therefore the probability that more than 3 will withdraw is 1 - 0.1071=0.8929

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.

5 0

4 years ago

0/1

pychu [463]

Answer:

68.8 sq. inches.

Step-by-step explanation:

The triangular surfaces have sides 2 in by 3 in by 4 in.

So, half perimeter, s = $\frac{2 + 3 + 4}{2} = 4.5$