Please explain to me how to do this or give me the answer.

Am not sure if I started this right, but it's probability, can someone check if I started right?

Sphinxa [80]

Yup! You got it so far! Keep it up! <3

6 0

3 years ago

i don't understand how to make it into a graphing form? how do you find the center and radius? (I'm confused because it has a 4x

Vilka [71]

Answer: $\bold{(x-2)^2+(y-3)^2=\dfrac{1}{4}}$

Center = (2, 3) radius = $\bold{\dfrac{1}{2}}$

<u>Step-by-step explanation:</u>

When both the x² and y² values are equal and positive, the shape is a circle. Complete the square to put the equation in format:

(x-h)² + (y-k)² = r² where

(h, k) is the vertex
r is the radius

1) Group the x's and y's together and move the number to the right side

4x² - 16x + 4y² - 24y = -51

2) Factor out the 4 from the x² and y²

4(x² - 4x ) + 4(y² - 6y ) = -51

3) Complete the square (divide the x and y value by 2 and square it)

$4[x^2-4x+\bigg(\dfrac{-4}{2}\bigg)^2]+4[y^2-6y+\bigg(\dfrac{-6}{2}\bigg)^2]=-51+4\bigg(\dfrac{-4}{2}\bigg)^2+4\bigg(\dfrac{-6}{2}\bigg)^2$

= 4(x - 2)² + 4(y - 3)² = -51 + 4(-2)² + 4(-3)²

= 4(x - 2)² + 4(y - 3)² = -51 + 4(4) + 4(9)

= 4(x - 2)² + 4(y - 3)² = -51 + 16 + 36

= 4(x - 2)² + 4(y - 3)² = 1

4) Divide both sides by 4

$(x-2)^2+(y-3)^2=\dfrac{1}{4}$

(h, k) = (2, 3)
$r=\dfrac{1}{2}$

6 0

3 years ago

An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

andrey2020 [161]

Answer:

The probability that A selects the first red ball is 0.5833.

Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are $E(1)= ^9C_2$ places in which the other 2 red balls can be placed.

A red ball drawn third, there are $E(3)= ^7C_2$ places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are $E(5)= ^5C_2$ places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are $E(7)= ^3C_2$ places in which the other 2 red balls can be placed.

The total number of total event is $S= ^{10}C_3$

The probability that A selects the first red ball is

$P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}$

$P(A \text{wins})=\frac{36+21+10+3}{120}$

$P(A \text{wins})=\frac{70}{120}$

$P(A \text{wins})=0.5833$

6 0

3 years ago

What functions have a range of {y £ R | -

Fudgin [204]

Answer:

the second and the fourth have a range of real no

4 0

3 years ago

Washington (CNN) About three-quarters of Americans (77%) approve of President Donald Trump's decision to meet with North Korean

snow_tiger [21]

Answer:

b. The margin of error would decrease.

Step-by-step explanation:

Margin of error of a confidence interval:

The margin of error of a confidence interval has the following format:

$M = z\frac{\sigma}{\sqrt{n}}$

In which z is related to the confidence level, $\sigma$ is the standard deviation of the population and n is the size of the sample.

This means that the margin of error is inversely proportional to the size of the sample, which means that if the sample size increases, the margin of error decreases.

In this question, the sample size is increased, leading to a smaller margin of error. So the correct answer is given by option b.

8 0

3 years ago