All categories

4 years ago

The old man walked 2/3 of a mile in 1/2 of an hour. What is this rate in miles per hour?

1 answer:

7 0

If the old man walks 2/3 of a mile in 1/2 an hour, you can find how many miles he walk in one hour by setting up a fraction where distance (miles) is the numerator and the denominator is time (hours). So, (2/3)/(1/2) is the fraction. Now, we have to figure out how to make the denominator one, and we do that by multiplying is by 2. Whatever you do to the denominator you have to do to the numerator, so multiply 2/3 by 2 and 1/2 by 2.

(1/2) x 2 = 1 and (2/3) x 2 = 4/3 or 1 1/3

the old man walked 1 1/3 miles per hour

You might be interested in

Consider the function f(x, y) = x2 + xy + y2 defined on the unit disc, namely, d = {(x, y)| x2 + y2 ≤ 1}. use the method of lagr

Pani-rosa [81]

First we note that the partial derivatives vanish simultaneously at one point:

$\begin{cases}f_x=2x+y=0\\f_y=x+2y=0\end{cases}\implies(x,y)=(0,0)$

so there is one critical point within the region $D$ .

The Lagrangian is

$L(x,y,\lambda)=x^2+xy+y^2+\lambda(x^2+y^2-1)$

and has partial derivatives

$L_x=2x+y+2\lambda x$
$L_y=x+2y+2\lambda y$
$L_\lambda=x^2+y^2-1$

Set each partial derivative to 0 to find the possible critical points within the disk $D$ . Then we notice that

$yL_x=2xy+y^2+2\lambda xy=0$
$xL_y=x^2+2xy+2\lambda xy=0$
$L_\lambda=0\implies x^2+y^2=1$

$\implies xL_y-yL_x=x^2-y^2=0$

Since $x^2+y^2=1$ , we have

$x^2-y^2=x^2+y^2-2y^2=0\implies1=2y^2\implies y^2=\dfrac12\implies y=\pm\dfrac1{\sqrt2}$

And since $x^2-y^2=0$ , or $x^2=y^2$ , we also have

$x=\pm\dfrac1{\sqrt2}$

So we have four possible additional critical points to consider:

$f(0,0)=0$
$f\left(-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac32$
$f\left(-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac12$
$f\left(\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac12$
$f\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac32$

It should be clear enough which of these correspond to the absolute extrema of $f$ over $D$ .

8 0

3 years ago

F(x) = x − 8, g(x) = |x + 8|<br> Find <br> (f ∘ g)(x).

miskamm [114]

Answer:

Step-by-step explanation:

$f(x) = x - 8 \\ \\ g(x) = |x + 8|$

To find (f ∘ g)(x) , substitute g(x) into f(x). That is for every x in f (x) replace it with g (x)

That's

Hope this helps you

5 0

4 years ago

Suppose there are 1,000 total sales representatives in the midwest and northeast regions, with 500 sales re

Maru [420]

Making 20 groups of 10 is the sampling strategy that can be used to select 200 sales representatives.

What is sampling?

Sampling is the phrase used to describe a statistical procedure in which a predetermined number of observations are chosen from a larger population. More precisely, sampling can be described as the process of choosing a portion of a statistical population in order to estimate the characteristics of the entire population. Samples that are representative of the population under study are sought for by statisticians.

Making 20 groups of 10 is the most effective way of sampling that can be employed here for the sales director to incorporate 200 sales representatives in this situation, with an equal amount of representatives from each region.

Learn more about sampling here:

brainly.com/question/11045407

#SPJ4

4 0

2 years ago

Mcd de 35 y 45? Ayuda por favor

aleksklad [387]

Espero que te sirva de ayuda.

8 0

3 years ago

Read 2 more answers

Let Y1 and Y2 have the joint probability density function given by:

Ann [662]

Answer:

a) k=6

b) P(Y1 ≤ 3/4, Y2 ≥ 1/2) = 9/16

Step-by-step explanation:

a) if

f (y1, y2) = k(1 − y2), 0 ≤ y1 ≤ y2 ≤ 1, 0, elsewhere

for f to be a probability density function , has to comply with the requirement that the sum of the probability of all the posible states is 1 , then

P(all possible values) = ∫∫f (y1, y2) dy1*dy2 = 1

then integrated between

y1 ≤ y2 ≤ 1 and 0 ≤ y1 ≤ 1

∫∫f (y1, y2) dy1*dy2 = ∫∫k(1 − y2) dy1*dy2 = k ∫ [(1-1²/2)- (y1-y1²/2)] dy1 = k ∫ (1/2-y1+y1²/2) dy1) = k[ (1/2* 1 - 1²/2 +1/2*1³/3)- (1/2* 0 - 0²/2 +1/2*0³/3)] = k*(1/6)

then

k/6 = 1 → k=6

P(Y1 ≤ 3/4, Y2 ≥ 1/2) = P (0 ≤Y1 ≤ 3/4, 1/2 ≤Y2 ≤ 1) = p

then

p = ∫∫f (y1, y2) dy1*dy2 = 6*∫∫(1 − y2) dy1*dy2 = 6*∫(1 − y2) *dy2 ∫dy1 =

6*[(1-1²/2)-((1/2) - (1/2)²/2)]*[3/4-0] = 6*(1/8)*(3/4)= 9/16

therefore

P(Y1 ≤ 3/4, Y2 ≥ 1/2) = 9/16

8 0

3 years ago