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2 years ago

GIVEAWAY!! QUICK QUESTION!!!

Mathematics

2 answers:

Dmitry [639]2 years ago

7 0

Answer:

-2

Step-by-step explanation:

Given :

An example of a coefficient ⇒ '11' in 11kk

Solving :

The coefficient is the part that precedes the variable(s)
Therefore, on solving -k - k
⇒ -2k
⇒ -2 is the coefficient

Olin [163]2 years ago

3 0

Answer:

-2

Step-by-step explanation:

Given expression: -k - k

Combine like terms and simplify:

⇒ -k - k

⇒ -(k + k)

⇒ -(2k) = -2k

Determining the constant:

The constant of the term MUST be an integer. It CANNOT be a variable. In this case, the only constant that is a cooeficient of k is -2.

Thus, the cooeficient of -k−k is -2.

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9. A ball is thrown into the air at an initial velocity of 18 meters per second from an initial height of 10

marusya05 [52]

D3 secaonds answer 20 meters=1

6 0

2 years ago

-4(x-1)=-3x+13-9-x PLS SOLVE IT STEP BY STEP

krek1111 [17]

-4(x-1)=-3x+13-9-x
-4x+4=-3x+4-x
-4x+4=-4x+4
-4x=-4x
0=0

4 0

2 years ago

Read 2 more answers

PLEASE HELP ASAP

Nat2105 [25]

A leg of a right triangle can never be greater then the hypotenuse

this formula will tell you if it's a valid right triangle
${?}^{2} + {?}^{2} = {?}^{2}$
the first two question marks are the legs of the triangle while the one after the = is the hypotenuse.

THE STEPS:
${3}^{2} + { \sqrt{7}^{2} } = {4}^{2}$
$9 + {2.65}^{2} = 16$
${2.65}^{2} = 7$
$9 + 7 = 16$
ANSWER: $\sqrt{7}$

5 0

2 years ago

The table shows the average annual cost of tuition at 4-year institutions from 2003 to 2010.

nata0808 [166]

Answer: 1) The best estimate for the average cost of tuition at a 4-year institution starting in 2020 =$ 31524.31

2) The slope of regression line b=937.97 represents the rate of change of average annual cost of tuition at 4-year institutions (y) from 2003 to 2010(x). Here,average annual cost of tuition at 4-year institutions is dependent on school years .

Step-by-step explanation:

1) For the given situation we need to find linear regression equation Y=a+bX for the given situation.

Let x be the number of years starting with 2003 to 2010.

i.e. n=8

and y be the average annual cost of tuition at 4-year institutions from 2003 to 2010.

With reference to table we get

$\sum x=36\\\sum y=150894\\\sum x^2=204\\\sum xy=718418$

By using above values find a and b for Y=a+bX, where b is the slope of regression line.

$a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}=\frac{150894(204)-(36)718418}{8(204)-(36)^2}=\frac{30782376-25863048}{1632-1296}=\frac{4919328}{336}\\\\=14640.85$

and

$b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}=\frac{8(718418)-(36)150894}{8(204)-(36)^2}=\frac{5747344-5432184}{1632-1296}=\frac{315160}{336}\\\\=937.97$

∴ To find average cost of tuition at a 4-year institution starting in 2020.(as n becomes 18 for year 2020 if starts from 2003 ⇒X=18)

So, Y= 14640.85 + 937.97×18 = 31524.31

∴The best estimate for the average cost of tuition at a 4-year institution starting in 2020 = $31524.31

4 0

3 years ago

Arm Span(x) Height(y)

natita [175]

Answer:

Here's what I get.

Step-by-step explanation:

1. Representation of data

I used Excel to create a scatterplot of the data, draw the line of best fit, and print the regression equation.

2. Line of best fit

(a) Variables

I chose arm span as the dependent variable (y-axis) and height as the independent variable (x-axis).

It seems to me that arm span depends on your height rather than the other way around.

(b) Regression equation

The calculation is easy but tedious, so I asked Excel to do it.

For the equation y = ax + b, the formulas are

$a = \dfrac{\sum y \sum x^{2} - \sum x \sumxy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\b = \dfrac{n\sumx y - \sum x \sumxy}{n\sum x^{2}- \left (\sum x\right )^{2}}$

This gave the regression equation:

y = 1.0595x - 4.1524

(c) Interpretation

The line shows how arm span depends on height.

The slope of the line says that arm span increases about 6 % faster than height.

The y-intercept is -4. If your height is zero, your arm length is -4 in (both are impossible).

(d) Residuals

$\begin{array}{cccr}&\textbf{Arm Span} & \textbf{Arm Span}&\\\textbf{Height/in} &\textbf{Actual} & \textbf{Predicted}&\textbf{Residual}\\25 & 19 & 22.3 & -3.3\\40 & 41 & 38.2 & 2.8\\55 & 51 & 54.1 & -3.1\\65 & 67 & 62.6 & 4.4\\ \end{array}$

The residuals appear to be evenly distributed above and below the predicted values.

A graph of all the residuals confirms this observation.

The equation usually predicts arm span to within 4 in.

(e) Predictions

(i) Height of person with 66 in arm span

$\begin{array}{rcl}y& = & 1.0595x - 4.1524\\66 & = & 1.0595x - 4.1524\\70.1524 & = & 1.0595x\\x & = & \dfrac{70.1524}{1.0595}\\\\& = & \textbf{66 in}\\\end{array}\\\text{A person with an arm span of 66 in should have a height of about $\large \boxed{\textbf{66 in}}$}$

(ii) Arm span of 74 in tall person

$\begin{array}{rcl}y& = & 1.0595x - 4.1524\\& = & 1.0595\times74 - 4.1524\\& = & 78.4030 - 4.1524\\& = & \textbf{74 in}\\\end{array}\\\text{ A person who is 74 in tall should have an arm span of $\large \boxed{\textbf{74 in}}$}$

6 0

2 years ago