Which term describes the set of all possible input values for a function?

Let´s call O the place for the gas station, A the location point of police car ( at 5 miles east from the gas station ), and B the location point of the truck ( 12 miles south from a gas station.

The three points shape a right triangle with L (distance between police car and truck) then

L² = OA² + OB² at the point police car is 5 miles east and truck 12 miles south ( both from gas station)

L² = ( 5)² + (12)²

L = √ 25 + 144

L = √169

L = 13 miles

Pitagoras theorem establishes in a right triangle hypotenuse L is:

L² = a² + b² a and b are the legs then

In general

L² = x² + y ² x and y the legs over distance police-car/gas station and truck/gas station

Differentiation on both sides of the equation with respect to time give us:

2*L*DL/dt = 2*x*Dx/dt + 2*y*Dy/dt

Where Dx/dt = 130 m/h West Dy/dt = 100 m/h south

L = 13 m x = 5 m y = 12

Then by substitution

2*13*DL/dt = 2*5*130 + 2*12*100

DL/dt = (1300+ 2400) / 26

DL/dt = 142 m/h

DL/dt is positive then L s increasing

b) If truck speed is 70 m/h

DL/dt still positive but with smaller module, then L still increasing but at smaller speed

DL/dt = 1300 + 2*12*70

DL/dt = (1300 + 1680)/ 26

DL/dt = 114,6 m

7 0

3 years ago

The results of Accounting Principals’ latest Workonomix survey indicate the average American worker spends $1092 on coffee annua

Nataly_w [17]

Answer:

a)

Group 18-34 years old

$\bar x = 1041.625 \\ s^2=485301 \\ s=696.635$

Group 35-44 years old

$\bar x = 1359.5 \\ s^2=178548 \\ s=422.549$

Group 45 and older

$\bar x = 1414.375 \\ s^2=18292.27 \\ s=135.248$

b)

According to the sample there is 9.04% probability that a person between 18 and 34 consume less than the average, 47.74% probability that a person between 35 and 44 consume more than the average and 50% probability that a person older than 45 consume more than the average.

Step-by-step explanation:

a)

The mean for each sample is

$\bar x=\frac{\sum_{k=1}^{10}x_k}{10}$

where the $x_k$ are the data corresponding to each group

The variance is

$s^2=\frac{\sum_{k=1}^{10}(\bar x-x_k)^2}{9}$

and the standard deviation is s, the square root of the variance.

Group 18-34 years old

$\bar x = 1041.625 \\ s^2=485301 \\ s=696.635$

Group 35-44 years old

$\bar x = 1359.5 \\ s^2=178548 \\ s=422.549$

Group 45 and older

$\bar x = 1414.375 \\ s^2=18292.27 \\ s=135.248$

b)

Let's compare these averages against the general media established of $1,092 by using the corresponding z-scores

$z=\frac{\bar x-\mu}{s/\sqrt{n}}$

where

$\bar x$ = mean of the sample

$\mu$ = established average

s = standard deviation of the sample

n = size of the sample

z-score of Group 18-34 years old

$z=\frac{1041.625-1092}{696.635/\sqrt{10}}=-0.2286$

The area under the normal curve N(0;1) between -0.2286 and 0 is 0.0904. So according to the sample there is 9.04% probability that a person between 18 and 34 consume less than the average.

z-score of Group 35-44 years old

$z=\frac{1359-1092}{422.5491/\sqrt{10}}=2.0019$

The area under the normal curve N(0;1) between 0 and 2.0019 is 0.4774. So according to the sample there is 47.74% probability that a person between 35 and 44 consume more than the average.

z-score of Group 45 and older

$z=\frac{1414.375-1092}{135.2489/\sqrt{10}}=7.5375$

The area under the normal curve N(0;1) between 0 and 7.5375 is 0.5. So according to the sample there is 50% probability that a person older than 45 consume more than the average.

3 0

4 years ago