53 dL = ______ daL im struggling with this one<br> .

Consider a nine -year moving average used to smooth a time series that was first recorded in 1961. a. Which year serves as the f

kirill [66]

Answer:

a. 1965

b. 8 years

Step-by-step explanation:

For answers to questions like these it can work to consider the smallest possible dataset.

<h3>Dataset</h3>

For our purpose, consider the 9 years/values to be averaged to be ...

1, 2, 3, 4, 5, 6, 7, 8 ,9

<h3>Observations</h3>

a. The center value of the data set is 5. Its number is 5-1=4 more than the first one.

The first centered value is from the year 1961 +4 = 1965.

b. The 4 values at the beginning, and the 4 values at the end do not have a corresponding "average" value. That is, 4+4 = 8 values in the series are lost with respect to the number of average values.

8 years of values are lost.

5 0

1 year ago

Science quiz has 12 multiple choice questions with four choices for each find the total number of ways to answer the questions

brilliants [131]

Answer:

Im guessing 48 ways cause 12 x 4 =48

6 0

2 years ago

Read 2 more answers

Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$

Simplify the radical with the cube:

$\displaystyle x^3=\frac{3y+16}{2y+9}$

Multiply by 2y+9

$\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)$

Simplify:

$\displaystyle x^3(2y+9)=3y+16$

Operate the parentheses:

$\displaystyle x^3(2y)+x^3(9)=3y+16$

$\displaystyle 2x^3y+9x^3=3y+16$

Subtract 3y and $9x^3$ :

$\displaystyle 2x^3y - 3y=16-9x^3$

Factor y out of the left side:

$\displaystyle y(2x^3 - 3)=16-9x^3$

Divide by $2x^3 - 3$ :

$\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}$

ii) To find y when x=2, substitute:

$\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}$

$\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}$

$\displaystyle y=\frac{16-72}{16- 3}$

$\displaystyle y=\frac{-56}{13}$

$\mathbf{\displaystyle y=-\frac{56}{13}}$

8 0

2 years ago

I will give brainliest. <br> What is the hypotenuse and how do I get it?

Ivan

Step-by-step explanation:

by Pythagoras theorem:

perpendicular² + base²= hypotenuse²

(1/√2)²+ (1/√3)² = H²

1/2+1/3= H²

5/6 = H²

√(5/6 )= H

√5/√6 = H

plz mark my answer as brainlist plzzzz.

hope this will be helpful to you

4 0

2 years ago

Which must be true of a quadratic function whose vertex is the same as the y-intercept ?

jeka94

A quadratic function whose vertex is the same as the y-intercept has the equation
y=x^2+k (where k is the y-intercept, with vertex (0,k))

Since the vertex coincides with the y-intercept, the axis of symmetry is x=0.

3 0

2 years ago

53 dL = ______ daL im struggling with this one .