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2 years ago

The difference between half of a number and 19.

1 answer:

6 0

The difference between two numbers is 19. Half the larger number plus twice the smaller number is 12. What are the numbers?

This is the equation I came up with but I can't make it work.
x-y=19
1/2x+2y=12

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Find the distance between the points 5 , 3 and -6 , 3 .

just olya [345]

(5, 3) (-6, 3)
The slope is 0 Slope = m = (Y2 -Y1) ÷ (X2 -X1) (Y2 -Y1)=0
SO, the distance is just the difference in the x values.
distance = 5 --6 = 11

7 0

3 years ago

Please help with the precalc question above...

Sonja [21]

Answer:

the last choice

Step-by-step explanation:

two functions are said to be inverse when they are symmetric about the line y=x

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2 years ago

Convert the following improper fractions into mixed numbers. 3 1/4

lara [203]

Answer:

7 and 3/4

Step-by-step explanation:

31/4=7-3

the remainder is 3/4

8 0

2 years ago

An artist sells original hand painted greeting cards. He makes $2.50 profit on a small card and $4.00 profit on a large card. He

Marysya12 [62]

Since he sells 5 large cards for every 2 small, we can multiply the number of small cards he needs to sell by 2.5 to find how many large cards he needs to sell.

So let x=the number of small cards sold, and 2.5x=the number of large cards sold

The equation is:Total profit=x+(2.5x)

Since a small card is $2.5, a large card is $4, and the total profit will be $10,000 plug those in to make the equation:

10,000=2.5x+4(2.5x)

We can combine the x's to get 10,000=12.5x and then divide both sides by 12.5 to get x=800.

So he needs to sell 800 small cards.

Multiply that by that 2.5.

And he needs to sell 2000 large cards.

3 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

3 years ago