2/7 times 5 and 1/2 simplified

Find the radius of the circle if the center is at (1, 2) and the point (-3, 4) lies on the circle.

Anton [14]

Answer:

2√5

Step-by-step explanation:

The radius is the distance from the center to a point anywhere along the edge of the circle. To find the distance between these two points, use the distance formula.

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(1--3)^2 + (2-4)^2}\\ d = \sqrt{4^2 + -2^2}\\ d = \sqrt{16 + 4}\\ d = \sqrt{20} \\d = 2\sqrt{5}$

7 0

3 years ago

Read 2 more answers

Solve the proportion using equivalent ratios. Explain the steps you used to solve the proportion, and include the answer in your

rodikova [14]

Answer: Your answer is X=6

Step-by-step explanation:

10/3 = 20/x

Determine the defined range

10/3 20/x, x=0

Cross-multiply

10x = 60

Divide both sides by 10

x=6,x=0

Check if the solution is in the defined range

SOLUTION x=6

I HOPE THIS HELPS IF NOT THE SORRY

4 0

3 years ago

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the n

Ivanshal [37]

Answer:

a) The probability that exactly 4 flights are on time is equal to 0.0313

b) The probability that at most 3 flights are on time is equal to 0.0293

c) The probability that at least 8 flights are on time is equal to 0.00586

Step-by-step explanation:

The question posted is incomplete. This is the complete question:

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures.

a) The probability that exactly 4 flights are on time is =

b) The probability that at most 3 flights are on time is =

c)The probability that at least 8 flights are on time is =

<h2>Solution to the problem</h2>

a) Probability that exactly 4 flights are on time

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

$P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}$

Where $_nC_x$ is the number of different combinations of x success in n trials.

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

Hence,

$P(X=4)=_9C_4\cdot (0.5)^4\cdot (0.5)^{5}$

$_9C_4=\frac{4!}{9!(9-4)!}=126$

$P(X=4)=126\cdot (0.5)^4\cdot (0.5)^{5}=0.03125$

b) Probability that at most 3 flights are on time

The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=0)=(0.5)^9=0.00195313$ . . . (the probability that all are not on time)

$P(X=1)=_9C_1(0.5)^1(0.5)^8=9(0.5)^1(0.5)^8=0.00390625$

$P(X=2)=_9C_2(0.5)^2(0.5)^7=36(0.5)^2(0.5)^7=0.0078125$

$P(X=3)= _9C_3(0.5)^3(0.5)^6=84(0.5)^3(0.5)^6=0.015625$

$P(X\leq 3)=0.00195313+0.00390625+0.0078125+0.015625=0.02929688\\\\ P(X\leq 3) \approx 0.0293$

c) Probability that at least 8 flights are on time 

That at least 8 flights are on time is the same that at most 1 is not on time.

That is, 1 or 0 flights are not on time.

Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.

$P(Y=0)=_0C_9(0.5)^0(0.5)^9=0.00195313\\ \\ P(Y=1)=_1C_9(0.5)^1(0.5)^8=0.0039065\\ \\ P(Y=0)+P(Y=1)=0.00585938\approx 0.00586$

6 0

3 years ago

Will give brainliest

Tamiku [17]

Answer:

D. promoting general friendliness

Step-by-step explanation:

You are not invading someone's personal life. You are just being friendly.

Hope this helps! Please mark as brainliest.

6 0

3 years ago

Read 2 more answers

Is the following correspondence a function

GenaCL600 [577]

No, because element b has 2 correspondents.

7 0

3 years ago

Read 2 more answers