Fill in the y values of the t-table

At what point does y = 3x -2 intercept :

bezimeni [28]

Y axis= -2
x axis= 2/3

5 0

3 years ago

If the coefficient of determination for a data set is 0.75 and the SSE for the data set is 9, what is the SST for the data set?

OverLord2011 [107]

0.1873 aaaaaaaaaaaaaaaaaaaaaaaa

6 0

3 years ago

Read 2 more answers

The area of sandbox in park is represented by 2X^2-5x-3 find the dimensions of the sandbox in terms of

lora16 [44]

Answer:

The dimension of the sandbox is (2x+1) by (x - 3)

Step-by-step explanation:

It seems the complete question will be:

The area of sandbox in park is represented by 2X^2-5x-3 find the dimensions of the sandbox in terms of x.

Step-by-step explanation:

From the question, the given expression is 2X^2-5x-3. This can be rewritten as

$2x^{2} -5x-3$

If the area of the sandbox in park is represented by this expression, then the dimensions of the sandbox will be the product of the factors. To determine the factors, we will factorize the given quadratic expression.

Factorizing the expression $2x^{2} -5x-3$ , we get

$2x^{2} -6x+x-3$

$2x(x-3)+1(x-3)$

$(2x+1)(x-3)$

Hence, the dimension of the sandbox is (2x+1) by (x - 3)

6 0

3 years ago

(\tan ^(2)\theta \cos ^(2)\theta -1)/(1+\cos (2\theta ))=

Vitek1552 [10]

(tan²(θ) cos²(θ) - 1) / (1 + cos(2θ))

Recall that

tan(θ) = sin(θ) / cos(θ)

so cos²(θ) cancels with the cos²(θ) in the tan²(θ) term:

(sin²(θ) - 1) / (1 + cos(2θ))

Recall the double angle identity for cosine,

cos(2θ) = 2 cos²(θ) - 1

so the 1 in the denominator also vanishes:

(sin²(θ) - 1) / (2 cos²(θ))

Recall the Pythagorean identity,

cos²(θ) + sin²(θ) = 1

which means

sin²(θ) - 1 = -cos²(θ):

-cos²(θ) / (2 cos²(θ))

Cancel the cos²(θ) terms to end up with

(tan²(θ) cos²(θ) - 1) / (1 + cos(2θ)) = -1/2

7 0

3 years ago

2log(2a - 1) = 0 How

Kryger [21]

Answer:

Step-by-step explanation:

$2\log(2a-1)=0\\\\\text{Domain:}\\\\2a-1>0\qquad\text{add 1 to both sides}\\2a-1+1>0+1\\2a>1\qquad\text{divide both sides by 2}\\\dfrac{2a}{2}>\dfrac{1}{2}\\\boxed{a>0.5}\\============================\\\\2\log(2a-1)=0\qquad\text{divide both sides by 2}\\\\\dfrac{2\!\!\!\!\diagup\log(2a-1)}{2\!\!\!\!\diagup}=\dfrac{0}{2}\\\\\log(2a-1)=0\qquad\text{use}\ \log_ab=c\iff b=a^c\\\\\log(2a-1)=\log10^0\\\\\log(2a-1)=\log1\iff2a-1=1\qquad\text{add 1 to both sides}\\\\2a-1+1=1+1\\\\2a=2\qquad\text{divide both sides by 2}\\\\\dfrac{2a}{2}=\dfrac{2}{2}\\\\a=1\in D$

3 0

3 years ago