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3 years ago

To prove a polygon is a rectangle which of the properties listed must be included in the proof

Mathematics

1 answer:

Zigmanuir [339]3 years ago

7 0

Answer:

if the diagonals of a parallelogram are congruent, then it's a rectangle (neither the reverse of the definition nor the converse of a property). If a parallelogram contains a right angle, then it's a rectangle (neither the reverse of the definition nor the converse of a property).

Step-by-step explanation:

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Please help me with this! Thank you

Iteru [2.4K]

Answer:

AA similarity theorm

Step-by-step explanation:

wx prependicular to xz and uz is same us wx

and if only the angles are given the figure would be similar with AA similarity theorm

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2 years ago

A surveyor found the angle of elevation from the ground to the top of the building at two locations 20 feet apart as shown below

Serga [27]

Applying the sine ratio and law of sines, the correct measurements are:

B. m∠B = 15°

E. h ≈ 31.28 ft.

<h3>What is the Sine Ratio?</h3>

Sine ratio that can be used to determine the side length of a right triangle is, sin ∅ = opposite side/hypotenuse.

Find c using the law of sines:

C = 33°

A = 180 - 48 = 132 [linear pair]

B = 180 - 33 - 132 = 15° [triangle sum theorem]

b = 20 ft

c = ?

Using the law of sines, b/sin B = c/sin C, we have:

20/sin 15 = c/sin 33

(c)(sin 15) = (20)(sin 33)

c = (20 × sin 33)/sin 15

c ≈ 42.09

Use the sine ratio to find h:

∅ = 48°

Hypotenuse = c = 42.09

Opposite = h = ?

sin 48 = h/42.09

h = (sin 48)(42.09)

h ≈ 31.28

The correct measurements are:

B. m∠B = 15°

E. h ≈ 31.28 ft

Learn more about the sine ratio on:

brainly.com/question/2920412

#SPJ1

4 0

1 year ago

A jar contains 5 black and 9 white marbles priscilla draws a marble from the jar what is the probability that she drew a black m

Vanyuwa [196]

The probability of the black marble will be 35.7% because 100/(9+5) is 7.14 and you multiply that by 5 to get 35.7%.

5 0

3 years ago

Read 2 more answers

Hi today im going to be giving a -lot of point so if u dont get points dont worry cuz im going to be doing 5 times.

oksian1 [2.3K]

Answer:

EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Thanks So MUCH!!!

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago