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3 years ago

Which of these conditions might be true if polygons ABCD and KLMN are similar?

Mathematics

1 answer:

White raven [17]3 years ago

8 0

Answer:

Option A.

Step-by-step explanation:

Properties of the similarity of two polygons,

1). All angles of one polygon will be equal to the angles of the other polygon.

2). Their corresponding sides will be in the same ratio.

If the measures of corresponding angles of KLMN and ABCD are equal, but the lengths of the corresponding sides of ABCD are half those of KLMN,

Then the polygons ABCD and KLMN will be similar.

Option A will be the answer.

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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

3 years ago

C= 2 7r; for r Plzz help

Nutka1998 [239]

Answer:

$C/27=r$