Ask question

Ask question

All categories

3 years ago

10

In March 1999, Bertrand Piccard and Brian Jones attempted to become the first people

1 answer:

lilavasa [31]3 years ago

3 0

Answer:

Distance traveled= 41968.4 km

The domain of the function is t

t(h)= 478

h = hours

Step-by-step explanation:

the distance that they traveled in kilometers, d, can be modeled by

d(t) = 97.8t

where t is the time in hours

They traveled a total of 478 hours.

Distance traveled d(t)

d(t) = 87.8(478)

d(t) = 41968.4

Distance traveled= 41968.4 km

The domain of the function is t

t(h)= 478

You might be interested in

Need help with this math question

anastassius [24]

Answer:

The vertex is: $(6, 8)$

Step-by-step explanation:

First solve the equation for the variable y

$x^2-4y-12x+68=0$

Add 4y on both sides of the equation

$4y=x^2-4y+4y-12x+68$

$4y=x^2-12x+68$

Notice that now the equation has the general form of a parabola

$ax^2 +bx +c$

In this case

$a=1\\b=-12\\c=68$

Add $(\frac{b}{2}) ^ 2$ and subtract $(\frac{b}{2}) ^ 2$ on the right side of the equation

$(\frac{b}{2}) ^ 2=(\frac{-12}{2}) ^ 2\\\\(\frac{b}{2}) ^ 2=(-6) ^ 2\\\\(\frac{b}{2}) ^ 2=36$

$4y=(x^2-12x+36)-36+68$

Factor the expression that is inside the parentheses

$4y=(x-6)^2+32$

Divide both sides of the equality between 4

$\frac{4}{4}y=\frac{1}{4}(x-6)^2+\frac{32}{4}$

$y=\frac{1}{4}(x-6)^2+8$

For an equation of the form

$y=a(x-h)^2 +k$

the vertex is: (h, k)

In this case

$h=6\\k =8$

the vertex is: $(6, 8)$

5 0

3 years ago

If the federal reserve sells $40,000 in treasury bonds to a bank at 5% interest, what is the immediate effect on the money suppl

lianna [129]

The money supply is decreased by $40,000

5 0

4 years ago

Read 2 more answers

Consider the given data. x 0 2 4 6 9 11 12 15 17 19 y 5 6 7 6 9 8 8 10 12 12 Use the least-squares regression to fit a straight

levacccp [35]

Answer:

See below

Step-by-step explanation:

By using the table 1 attached (See Table 1 attached)

We can perform all the calculations to express both, y as a function of x or x as a function of y.

Let's make first the line relating y as a function of x.

<u>y as a function of x </u>

<em>(y=response variable, x=explanatory variable) </em>

$\bf y=m_{yx}x+b_{yx}$

where

$\bf m_{yx}$ is the slope of the line

$\bf b_{yx}$ is the y-intercept

In this case we use these formulas:

$\bf m_{yx}=\frac{(\sum y)(\sum x)^2-(\sum x)(\sum xy)}{n\sum x^2-(\sum x)^2}$

$\bf b_{yx}=\frac{n\sum xy-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}$

n = 10 is the number of observations taken (pairs x,y)

<u>Note:</u> <em>Be careful not to confuse </em>

$\bf \sum x^2$ with $\bf (\sum x)^2$

Performing our calculations we get:

$\bf m_{yx}=\frac{(83)(95)^2-(95)(923)}{10*1277-(95)^2}=176.6061$

$\bf b_{yx}=\frac{10*923-(95)(83)}{10(1277)-(95)^2}=0.3591$

So the equation of the line that relates y as a function of x is

<h3>y = 176.6061x + 0.3591 </h3>

In order to compute the standard error $\bf S_{yx}$ , we must use Table 2 (See Table 2 attached) and use the definition

$\bf s_{yx}=\sqrt{\frac{(y-y_{est})^2}{n}}$

and we have that standard error when y is a function of x is

$\bf s_{yx}=\sqrt{\frac{39515985}{10}}=1987.8628$

Now, to find the line that relates x as a function of y, we simply switch the roles of x and y in the formulas.

So now we have:

x as a function of y

(x=response variable, y=explanatory variable)

$\bf x=m_{xy}y+b_{xy}$

where

$\bf m_{xy}$ is the slope of the line

$\bf b_{xy}$ is the x-intercept

In this case we use these formulas:

$\bf m_{xy}=\frac{(\sum x)(\sum y)^2-(\sum y)(\sum xy)}{n\sum y^2-(\sum y)^2}$

$\bf b_{xy}=\frac{n\sum xy-(\sum x)(\sum y)}{n(\sum y^2)-(\sum y)^2}$

n = 10 is the number of observations taken (pairs x,y)

<u>Note:</u> <em>Be careful not to confuse </em>

$\bf \sum y^2$ with $\bf (\sum y)^2$

Remark:<em> </em><em>If you wanted to draw this line in the classical style (the independent variable on the horizontal axis), you would have to swap the axis X and Y) </em>

Computing our values, we get

$\bf m_{xy}=\frac{(95)(83)^2-(83)(923)}{10*743-(83)^2}=1068.1072$

$\bf b_{xy}=\frac{10*923-(95)(83)}{10(743)-(83)^2}=2.4861$

and the line that relates x as a function of y is

<h3>x = 1068.1072y + 2.4861 </h3>

To find the standard error $\bf S_{xy}$ we use Table 3 (See Table 3 attached) and the formula

$\bf s_{xy}=\sqrt{\frac{(x-x_{est})^2}{n}}$

and we have that standard error when y is a function of x is

$\bf s_{xy}=\sqrt{\frac{846507757}{10}}=9200.5856$

<em>In both cases the correlation coefficient r is the same and it can be computed with the formula: </em>

$\bf r=\frac{\sum xy}{\sqrt{(\sum x^2)(\sum y^2)}}$

Remark: <em>This formula for r is only true if we assume the correlation is linear. The formula does not hold for other kind of correlations like parabolic, exponential,..., etc. </em>

Computing the correlation coefficient :

$\bf r=\frac{923}{\sqrt{(1277)(743)}}=0.9478$

5 0

4 years ago

write an equation in point slope form that represents a line that passes by (-9,-2) and has a slope of -4

KiRa [710]

Point slope: y-y1 = m(x-x1)
Plug in the numbers
Solution: y+2 = -4(x + 9)

8 0

3 years ago

Read 2 more answers

Cuales el numero menor que 34.969.387 y mayor que 34.969.385

gayaneshka [121]

Has la serie numérica

a= 34.969.385

b = 34.969.386

c = 34.969.387

$a < b < c$

8 0

4 years ago