The frist three terms of a geometric sequence are as follows 80,40,20

HELP PLEASE WILL GIVE BRAINLIEST WORTH 50 POINTS

Alja [10]

Answer:

y = 8.13 x -16125.713

hope it helps

Step-by-step explanation:

Sum of X = 14004

Sum of Y = 977

Mean X = 2000.5714

Mean Y = 139.5714

Sum of squares (SSX) = 53.7143

Sum of products (SP) = 436.7143

Regression Equation = ŷ = bX + a

b = SP/SSX = 436.71/53.71 = 8.13032

a = MY - bMX = 139.57 - (8.13*2000.57) = -16125.71277

ŷ = 8.13032X - 16125.71277

8 0

3 years ago

Read 2 more answers

(02.09) Given a triangle with vertices at points (4, 4), (4, 1), and (2, 1), which set of ordered pairs represents the new verti

wel

Hello!

This triangle is currently located in Quadrant 1. If it is rotated 90 degrees counter-clockwise around the origin, it will end up in Quadrant 2. The formula for this is seen below.

(x,y)⇒(-y,x)

We will do this to each point and get our final points.

(4,4)⇒(-4,4)
(4,1)⇒(-1,4)
(2,1)⇒(-1,2)

Therefore, the final points of our rotated triangle are (-4,4), (-1,4) and (-1,2).

I hope this helps!

4 0

3 years ago

How much will $5000 be worth in 5 years if it is compounded continuously at 3% interest?

Finger [1]

$5,796 is how much it would be. hope it helps

5 0

3 years ago

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Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$