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Zinaida [17]
3 years ago
11

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

Mathematics
1 answer:
murzikaleks [220]3 years ago
4 0
I only drink juice for breakfast time. I think that's the answer
You might be interested in
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:

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

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

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

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

<em>The distance of two points formula</em>

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}

<em>The diameter 'd' = 2 r</em>

                       2 r = √80

                            = \sqrt{16 X 5}

                           = 4 \sqrt{5}

                     <em> r =  2√5</em>

<em>Mid-point of two end points </em>

<em>                        </em>(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )<em></em>

<em>                                               = (-1 ,2)</em>

<em>Mid-point of two end points = center of the circle</em>

<em>                                     (h,k) = (-1 , 2)</em>

                 

The equation of the circle

           (x -h )² +(y-k)² = r²

<em>          (x -(-1) )² +(y-(2))² = (2(√5))²</em>

<em>         x² + 2 x + 1 + y² - 4 y + 4 = 20</em>

<em>          x² + 2 x + y² - 4 y  = 20 -5</em>

<em>           x² + 2 x + y² - 4 y  = 15</em>

<u><em>Final answer</em></u><em>:-</em>

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

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<em>                  </em>

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 <br><br> 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
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