All categories

3 years ago

The midpoint of JKis M (6, 3). One endpoint is J (14, 9). Find the coordinates of endpoint K.

Mathematics

2 answers:

Oksanka [162]3 years ago

7 0

Answer:

The coordinates of endpoint K are (-2,-3)

Step-by-step explanation:

Midpoint

Given two points J(jx, jy) and K(kx, ky), the coordinates of the midpoint M(mx,my) are given by:

$\displaystyle m_x=\frac{j_x+k_x}{2}$

$\displaystyle m_y=\frac{j_y+k_y}{2}$

We are given the coordinates of M and need to find the coordinates of K. From the equations above, solve for kx and ky:

$k_x=2m_x-j_x$

$k_y=2m_y-j_y$

The points are M(6,3) and J(14,9), thus:

$k_x=2\cdot 6-14=-2$

$k_y=2\cdot 3-9=-3$

The coordinates of endpoint K are (-2,-3)

Jobisdone [24]3 years ago

4 0

The endpoint K (-2, -3)

You might be interested in

Past records indicate that the probability of online retail orders

Tcecarenko [31]

Answer:

a) Mean = 1.6, standard deviation = 1.21

b) 18.87% probability that zero online retail orders will turn out to be fraudulent.

c) 32.82% probability that one online retail order will turn out to be fraudulent.

d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.

Step-by-step explanation:

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

The mean of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$p = 0.08, n = 20$

a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?

Mean

$E(X) = np = 20*0.08 = 1.6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21$

b. What is the probability that zero online retail orders will turn out to be fraudulent?

This is P(X = 0).

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

$P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887$

18.87% probability that zero online retail orders will turn out to be fraudulent.

c. What is the probability that one online retail order will turn out to be fraudulent?

This is P(X = 1).

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

$P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282$

32.82% probability that one online retail order will turn out to be fraudulent.

d. What is the probability that two or more online retail orders will turn out to be fraudulent?

Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So

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

We want $P(X \geq 2)$

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

In which

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

From itens b and c

$P(X \leq 1) = 0.1887 + 0.3282 = 0.5169$

$P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831$

48.31% probability that two or more online retail orders will turn out to be fraudulent.

4 0

4 years ago

You build and sell toy cars for $36 each. So far, you have built 25 toy cars. How many more toy cars will you have to build and

chubhunter [2.5K]

If you have built 25 cars, and they are 36 dollars per, then multiply 25 and 36 for the amount of money. This will result with 25*36=900. Since we already have 900 dollars worth, we subtract that from the goal of 1620, leaving us with 1620-900=720. Now, divide this by the price per car for the amount of cars needed to get to this goal. 720/36=20 cars

Hope this helps!

6 0

3 years ago

Read 2 more answers

What is the place value of the digit 9 in the number 69143

marshall27 [118]

It is the thousands place

7 0

3 years ago

Read 2 more answers

The area of a square field is 14400 sq. metre. Find the length of the side of the square.

Artist 52 [7]

Answer: