Which linear function represents the table?

If 12 gallon of paint covers 18 of a wall, then how many quarts of paint are needed for the entire wall?

marusya05 [52]

Answer: 16 quarts

Step-by-step explanation:

There are 4 quarts per 1 gallon.

You need 4 gallons total to cover the wall.

4 x 4=16

4 0

3 years ago

Sam's mother is twice as old as sam, In 12 years, Sam will be as old as his mother was 12 years ago. How old are is sam and his

irinina [24]

Answer:

12 and 24

Step-by-step explanation:

cuz if he's 12 and she's 24 in 12 years he'll be 24

4 0

3 years ago

Read 2 more answers

What is the solution set represented by this number line graph

Maru [420]

Answer:

x ≥ 2

Step-by-step explanation:

The dot is shaded on the point positive 2 and the arrow is going right so its x ≥ 2.

6 0

3 years ago

The Eco Pulse survey from the marketing communications firm Shelton Group asked individuals to indicate things they do that make

Bad White [126]

Answer:

a) There is a probability of 42% that the person will feel guilty for only one of those things.

b)There is a probability of 46% that a randomly selected person will not feel guilty for either of these reasons

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, for those people that will feel guilty about wasting food.

-Set B, for those people that will feel guilty about leaving lights on when not in a room.

The most important information is that there is a .12 probability that a randomly selected person will feel guilty for both of these reasons. It means that $P(A \cap B) = .12.$

The problem also states that there is a .39 probability that a randomly selected person will feel guilty about wasting food. It means that P(A) = 0.39. The probability of a person feeling guilty for only wasting food is PO(A) = .39-.12 = .27.

Also, there is a .27 probability that a randomly selected person will feel guilty about leaving lights on when not in a room. So, the probability of a person feeling guilty for only leaving the lights on is PO(B) = 0.27-0.12 = 0.15.

a) What is the probability that a randomly selected person will feel guilty for either wasting food or leaving lights on when not in a room?

This is the probability that the person feels guilt for only one of those things, so:

P = PO(A) + PO(B) = 0.27 + 0.15 = 0.42 = 42%

b) What is the probability that a randomly selected person will not feel guilty for either of these reasons

The sum of all the probabilities is always 1. In this problem, we have the following probabilies

- The person will not feel guilty for either of these reasons: P

- The person will feel guilty for only one of those things: PO(A) + PO(B) = 0.42

- The person will feel guilty for both reasons: PB = 0.12

So

`P + 0.42 + 0.12 = 1

P = 1-0.54

P = 0.46

There is a probability of 46% that a randomly selected person will not feel guilty for either of these reasons

4 0

3 years ago

Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

leonid [27]

Let $P=(x,y,z)$ be an arbitrary point on the surface. The distance between $P$ and the given point $(7,11,0)$ is given by the function

$d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}$

Note that $f(x)$ and $f(x)^2$ attain their extrema, if they have any, at the same values of $x$ . This allows us to consider the modified distance function,

$d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2$

So now you're minimizing $d^*(x,y,z)$ subject to the constraint $z^2=xy$ . This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

$\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)$

which has partial derivatives

$\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}$

Setting all four equation equal to 0, you find from the third equation that either $z=0$ or $\lambda=-1$ . In the first case, you arrive at a possible critical point of $(0,0,0)$ . In the second, plugging $\lambda=-1$ into the first two equations gives

$\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10$

and plugging these into the last equation gives

$z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5$

So you have three potential points to check: $(0,0,0)$ , $(2,10,2\sqrt5)$ , and $(2,10,-2\sqrt5)$ . Evaluating either distance function (I use $d^*$ ), you find that

$d^*(0,0,0)=170$
$d^*(2,10,2\sqrt5)=46$
$d^*(2,10,-2\sqrt5)=46$

So the two points on the surface $z^2=xy$ closest to the point $(7,11,0)$ are $(2,10,\pm2\sqrt5)$ .

5 0

4 years ago