What are two conditional statements that make up “ I dont drink juice if and only if it is breakfast time.”

If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle

stira [4]

Answer:

The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²

 or 

The equation of the circle x² + 2 x + y² - 4 y = 15

Step-by-step explanation:

Given points end Points are p(-3,-2) and q( 1,6)

The distance of two points formula

P Q = $\sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2} }$

P Q = $\sqrt{1 - (-3)^{2} + ((6 -(-2))^{2} }$

P Q = $\sqrt{16+64} = \sqrt{80}$

The diameter 'd' = 2 r

2 r = √80

= $\sqrt{16 X 5}$

= $4 \sqrt{5}$

r = 2√5

Mid-point of two end points

$(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )$

= (-1 ,2)

Mid-point of two end points = center of the circle

(h,k) = (-1 , 2)

The equation of the circle

(x -h )² +(y-k)² = r²

(x -(-1) )² +(y-(2))² = (2(√5))²

x² + 2 x + 1 + y² - 4 y + 4 = 20

x² + 2 x + y² - 4 y = 20 -5

x² + 2 x + y² - 4 y = 15

Final answer:-

The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²

 or 

The equation of the circle x² + 2 x + y² - 4 y = 15

6 0

3 years ago

How do you solve 2/7m -1/7 = 3/14

Mars2501 [29]

the only thing you need to do is separate m:

2/7m= 3/14 + 1/7

2/7m= 5/14

m= (5/14)/(2/7)

m= 5/14 × 7/2

m= 35/28

m= 5/4

4 0

2 years ago

Read 2 more answers

Let X1 and X2 be independent random variables with mean μand variance σ².

My name is Ann [436]

Answer:

a) $E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

b) $Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

Step-by-step explanation:

For this case we know that we have two random variables:

$X_1 , X_2$ both with mean $\mu = \mu$ and variance $\sigma^2$

And we define the following estimators:

$\hat \theta_1 = \frac{X_1 + X_2}{2}$

$\hat \theta_2 = \frac{X_1 + 3X_2}{4}$

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

$E(\hat \theta_i) = \mu , i = 1,2$

So let's find the expected values for each estimator:

$E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})$

Using properties of expected value we have this:

$E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

For the second estimator we have:

$E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})$

Using properties of expected value we have this:

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

$Var(aX) = a^2 Var(X)$

For the first estimator we have:

$Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})$

$Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

For the second estimator we have:

$Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})$

$Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

7 0

3 years ago

75 times 243 If your doing just for points I’ll report you

diamong [38]

18,225 :) have a nice day!

3 0

3 years ago

Read 2 more answers

A rectangular room twice as long as it is wide would contain 160 square feet more if each side were one foot longer. what is the

Blababa [14]

Length= 2*width

area=l*w=2*width*width=2*width^2
area+160=(2*width+1)*(width+1)
A+160=2w^2+2w+w+1
A+160=2w^2+3w+1
putting value of a
2w^2+160=2w^2+3w+1
160=3w+1
w=53m
l=106m

7 0

4 years ago