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Mnenie [13.5K]
2 years ago
10

Complete the similarity statement for the two triangles shown. Enter your answer in the box. △XBR∼△ Two similar triangles B R X

and N J Y. In triangle B R X, side B X is the base. Side B R is labeled 30 centimeters, side R X is labeled 40 centimeters, and side B X is labeled 60 centimeters. In triangle N J Y, side N Y is the base. Side N J is labeled 15 centimeters, side J Y is labeled 20 centimeters, and side N Y is labeled 30 centimeters. The measure of angle B equals the measure of angle N, the measure of angle R equals the measure of angle J, and the measure of angle X equals the measure of angle Y.

Mathematics
1 answer:
Andreyy892 years ago
5 0

Answer:  

Here, BRX and NJY are two triangles in which,

BR = 30 cm, RX = 40 cm, BX = 60 cm, NJ = 15 cm, JY = 20 cm and NY = 30 cm,

Also, m∠B = m∠N, m∠R = m∠J and m∠X = m∠Y,

By the property of congruence,

\angle B \cong \angle N, \angle R \cong \angle J and \angle X \cong \angle Y

Thus, By AAA similarity postulate,

\triangle BRX\sim \triangle NJY

Hence, proved.

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A class with n kids lines up for recess. The order in which the kids line up is random with eachordering being equally likely. T
Elis [28]

Answer:

A) P(Betty is first in line and mary is last) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))

B) The method used is Relative frequency approach.

Step-by-step explanation:

From the question, we are told a sample of n kids line up for recess.

Now, the order in which they line up is random with each ordering being equally likely. Thus, this means that the probability of each kid to take a position is n(total of kids/positions).

Since we are being asked about 3 kids from the class, let's assign a letter to each kid:

J: John

B: Betty

M: Mary

A) Now, we want to find the probability that Betty is first in line or Mary is last in line.

In this case, the events are not mutually exclusive, since it's possible that "Betty is first but Mary is not last" or "Mary is last but Betty is not first" or "Betty is the first in line and Mary is last". Thus, there is an intersection between them and the probability is symbolized as;

P(B₁ ∪ Mₙ) = P(B₁) + P (Mₙ) - P(B₁ ∩ Mₙ) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))

Where;

The suffix 1 refers to the first position while the suffix n refers to the last position.

Also, P(B₁ ∩ Mₙ) = P(B₁) × P(Mₙ/B₁)

This is because the events "Betty" and "Mary" are not independent since every time a kid takes his place the probability of the next one is affected.

B) The method used is Relative frequency approach.

In this method, the probabilities are usually assigned on the basis of experimentation or historical data.

For example, If A is an event we are considering, and we assume that we have performed the same experiment n times so that n is the number of times A could have occurred.

Also, let n_A be the number of times that A did occur.

Now, the relative frequency would be written as (n_A)/n.

Thus, in this method, we will define P(A) as:

P(A) = lim:n→∞[(n_A)/n]

7 0
2 years ago
isco Fever is randomly found in one half of one percent of the general population. Testing a swatch of clothing for the presence
maks197457 [2]

Answer:

0.8894 is the probability that the test result comes back negative if the disease is present .

Step-by-step explanation:

We are given the following in the question:

P(Disco Fever) = P( Disease) =

P(D) = \dfrac{1}{2}\times 1\% = 0.5 \times 0.01 = 0.005

Thus, we can write:

P(No  Disease) =

P(ND) =1 - P(D)= 0.995

P(Test Positive given the presence of the disease) = 0.99

P(TP | D ) = 0.99

P( false-positive) = 4%

P( TP | ND) = 0.04

We have to evaluate the probability that the test result comes back negative if the disease is present, that is

P(test result comes back negative if the disease is present)

By Bayes's theorem, we can write:

P(ND|TP) = \dfrac{P(ND)P(TP|ND)}{P(ND)P(TP|ND) + P(D)P(TP|D)}\\\\P(ND|TP) = \dfrac{0.995(0.04)}{0.995(0.04) + 0.005(0.99)} = 0.8894

0.8894 is the probability that the test result comes back negative if the disease is present .

5 0
1 year ago
Marika is training for a race track. she starts by sprinting 100 yards. she gradually increases her distance, adding 4 yards a d
iren2701 [21]

Answer: the correct answer is D.

Step-by-step explanation:

4 0
1 year ago
We know that a triangle with side lengths x2 -1, 2x, and x2 1 is a right triangle. Using those side lengths, find the missing tr
Olenka [21]

Least to greatest, the triples are (2x, x²-1, x²+1). Thus the value in the first column of the table is half the value of the first of the triples. Your completed table will be

\begin{array}{cc}\text{x-value}&\text{triple}\\3&\bf{(6,8,10)}\\\bf{4}&(8,15,17)\\5&\bf{(10,24,26)}\\\bf{6}&(12,35,36)\end{array}

4 0
1 year ago
Randy presses RAND on his calculator twice to obtain two random numbers between 0 and 1. Let $p$ be the probability that these t
anygoal [31]

Answer:

\frac{\pi}{4}

Step-by-step explanation:

Lets call x,y the numbers we obtain from the calculator. x and y are independent random variables of uniform[0,1] distribution.

Lets note that, since both x and y are between 0 and 1, then 1 is the biggest side of the triangle.

Lets first make a geometric interpretation. If the triangle were to be rectangle, then the side of lenght 1 should be its hypotenuse, and therefore x and y should satisfy this property:

x²+y² = 1

Remember that in this case we are supposing the triangle to be rectangle. But the exercise asks us to obtain an obtuse triangle. For that we will need to increase the angle obtained by the sides of lenght x and y. We can do that by 'expanding' the triangle, but if we do that preserving the values of x and y, then the side of lenght 1 should increase its lenght, which we dont want to. Thus, if we expand the triangle then we should also reduce the value of x and/or y so that the side of lenght 1 could preserve its lenght. With this intuition we could deduce that

x²+y² < 1

Now lets do a more mathematical approach. According to the Cosine theorem, a triangle of three sides of lenght a,b,c satisfies

a² = b²+c² - 2bc* cos(α), where α is the angle between the sides of lenght b and c.

Aplying this formula to our triangle, we have that

1^2 = x^2 + y^2 - 2bc* cos(\alpha) , where \alpha is the angle between the sides of lenght x and y.

Since the triangle is obtuse, then \pi/2 <  \alpha < \pi , and for those values cos(\alpha) is negative , hence we also obtain

1 > x² + y²

Thus, we need to calculate P(x²+y² <1). This probability can be calculated throught integration. We need to use polar coordinates.

(x, y) = (r*cosФ,r*senФ)

Where r is between 0 and 1, and Ф is between 0 and \pi /2 (that way, the numbers are positive).

The jacobian matrix has determinant r, therefore,

{\int\int}\limits_{x^2+y^2 < 1}  \, dxdy = \int\limits^1_0\int\limits^{\frac{\pi}{2}}_0 {r} \, d\phi dr = \frac{\pi}{2} * \int\limits^1_0 {r} \, dr =    \frac{\pi}{2} * (\frac{r^2}{2} |^1_0) = \frac{\pi}{4}

As a conclusion, the probability of obtaining an obtuse triangle is \frac{\pi}{4} .

6 0
1 year ago
Read 2 more answers
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