What equation results from completing the square and then factoring?

PLEASE HELP FAST WILL MARK BRAINLIEST PLZ

Jobisdone [24]

(-4, 3, 10). I'm not really sure tho

3 0

3 years ago

11. In circle H, if m3DHG = (11x - 36)* and m3GHF = (x + 12), find mDG.

Crazy boy [7]

Answer:

B. 187

Step-by-step explanation:

I think this is correct I took a test on this before

3 0

4 years ago

Read 2 more answers

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp

motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\$

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305$

The degrees of freedom for this sample size are:

$df=n-1=50-1=49$

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.01 \cdot 0.305=0.61$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95$

The 95% confidence interval for the mean is (5.73, 6.95).

4 0

4 years ago

We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independent

Bogdan [553]

Answer:

(a) P(X = 0) = 1/3

(b) P(X = 1) = 2/9

(c) P(X = −2) = 1/9

(d) P(X = 3) = 0

(a) P(Y = 0) = 0

(b) P(Y = 1) = 1/3

(c) P(Y = 2) = 1/3

Step-by-step explanation:

Given:

- Two 3-sided fair die.

- Random Variable X_1 denotes the number you get for rolling 1st die.

- Random Variable X_2 denotes the number you get for rolling 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

- First we will develop a probability distribution of X such that it is defined by the difference of second and first roll of die.

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

- The corresponding probabilities for each outcome are:

( X = -2 ): { X_2 = 1 , X_1 = 3 }

P ( X = -2 ): P ( X_2 = 1 ) * P ( X_1 = 3 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

( X = -1 ): { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }

P ( X = -1 ): P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 0 ): { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } + { X_2 = 3 , X_1 = 3 }

P ( X = -1 ):P ( X_2 = 1 )*P ( X_1 = 1 )+P( X_2 = 2 )*P ( X_1 = 2)+P( X_2 = 3 )*P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 3 / 9 ) = ( 1 / 3 )

( X = 1 ): { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }

P ( X = 1 ): P ( X_2 = 2 ) * P ( X_1 = 1 ) + P ( X_2 = 3 ) * P ( X_1 = 2)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 2 ): { X_2 = 1 , X_1 = 3 }

P ( X = 2 ): P ( X_2 = 3 ) * P ( X_1 = 1 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

- The distribution Y = X_2,

P(Y=0) = 0

P(Y=1) = 1/3

P(Y=2) = 1/ 3

- The probability for each number of 3 sided die is same = 1 / 3.

7 0

3 years ago

Use the power-reducing formulas to rewrite the expression as an equivalent expression that does not contain powers of trigonome

jolli1 [7]

Answer:

Answer is attached

8 0

3 years ago