Examine the work shown. Explain the error and find the correct result. 2(4 - 16) - (-30)2-12) - (-30)
24 - (-30)
54

Answers

Answer 1

Answer:

Answer:

-54

Step-by-step explanation:

When you solve any equation always remember to use PEMDAS (Parentheses Exponents, Multiplication, Division, Addition and Subtraction) use also move left to right. The correct way to solve this is

2(4-16)-30

2(-12)-30

-24-30

-54

Hope this helped

Answer 2

Answer:

Answer:

Step-by-step explanation:

Third line. "24 - (-30)" has to multiply 2 with -12, so it would be "-24" and not "24".

2(4 - 16) - (-30)

2(-12) + 30

-24 + 30 = 6

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Let Upper C left-parenthesis q right-parenthesis represent the cost, Upper R left-parenthesis q right-parenthesis the revenue, and pi left-parenthesis q right-parenthesis the total profit, in dollars, of producing q items.(a) If Upper C prime left-parenthesis 50 right-parenthesis equals 75 and Upper R prime left-parenthesis 50 right-parenthesis equals 88, approximately how much profit is earned by the 51 Superscript st item?The profit earned from the 51 Superscript st item will be approximately_______ dollar-sign.(b) If Upper C prime left-parenthesis 90 right-parenthesis equals 71 and Upper R prime left-parenthesis 90 right-parenthesis equals 67, approximately how much profit is earned by the 91 Superscript st item?The profit earned from the 91 Superscript st item will be approximately______ dollar-sign
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Find the scale factor of the line segment dilation. AB: endpoints (-6, -3) and (-3,-9) to A'B': endpoints at (-2, -1) and (-1, -3). A) -1/3 B) 1/3 C) 3D) -3

Answers

we know that the endpoints of AB are

$\begin{gathered} (x_1,y_1)=\mleft(-6,-3\mright) \n (x_2,y_2)=(-3,-9) \end{gathered}$

and the distance formula is given by

$d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$

By substituying these points, we have that

$d=\sqrt[]{(-3-(-6))^2+(-9-(-3))^2}$

which is equal to

$\begin{gathered} d=\sqrt[]{(-3+6)^2+(-9+3)^2} \n d=\sqrt[]{3^2+(-6)^2} \end{gathered}$

then

$\begin{gathered} d=\sqrt[]{9+36} \n d=\sqrt[]{45} \n d=\sqrt[]{9\cdot5} \n d=\sqrt[]{9}\cdot\sqrt[]{5} \n d=3\sqrt[]{5}\ldots..(A) \end{gathered}$

On the other hand, if

$\begin{gathered} (x_1,y_1)=(-2,-1) \n (x_2,y_2)=(-1,-3) \end{gathered}$

similarly to the previous case, the distance between the endpoint for A'B' is

$d=\sqrt[]{(-1-(-2))^2+(-3-(-1))^2}$

which is equal to

$\begin{gathered} d=\sqrt[]{(-1+2)^2+(-3+1)^2} \n d=\sqrt[]{1^2+(-2)^2} \n d=\sqrt[]{1+4} \n d=\sqrt[]{5}\ldots..(B) \end{gathered}$

Now, by comparing equation A and equation B, we can see that, the scale factor is 1/3.

Then, the answer is B.

For every 2 books that Tom buys, he can buy 3 magazines. If he pays $60 fo 4 books and 4 magazines, how much does he have to pay for each book?

Answers

Answer:

Price of book is $9\$$

Step-by-step explanation:

Let the price of book be $x$ and price of magazine be $y$

We have

$2x=3y$

We also have

$4x+4y=60$

$2* 2x+4y=60\n\n2*3y+4y=60\n\n10y=60\n\ny=6\n\n2x=3*6\n\nx=9\$$

Price of book is $9\$$

A spinner has 3 equal sections that are purple, green, and blue. What is the probability of not landing on blue?

Answers

Answer:

P(not landing on blue section)=2/3

Step-by-step explanation:

Not landing on blue section = landing on one of the other sections

the remaining sections are 2 each one has a P=1/3

2 × 1/3 = 2/3

Answer:

66%

Step-by-step explanation:

2 divided by 3= 0.666666666 which is then converted to percent (66%)

HOSWhich sentence would make the best cliff-hanger for the end of chapter 1?
The air was very cold.
In the distance, firelight glowed.
He bumped his foot on the side of the boat.
The men were very hungry.

Answers

Answer:

Step-by-step explanation:

Answer:

It's B. In the distance, firelight glowed.

Step-by-step explanation:

♥ right on edge ♥

19. Round the number 347 500 to the nearest 1000; 10 000 and 100 000.To the nearest 1000
To the nearest 10 000
To the nearest 100 000

Answers

Answer:

34800, 350000, 300000

Step-by-step explanation:

5>=5

34800

7>=5

350000

4<=5

300000

Three friends — let’s call them X, Y , and Z — like to play pool (pocket billiards). There are some pool games that involve three players, but these people instead like to play 9-ball, which is a game between two players with the property that a tie cannot occur (there’s always a winner and a loser in any given round). Since it’s not possible for all three of these friends to play at the same time, they use a simple rule to decide who plays in the next round: loser sits down. For example, suppose that, in round 1, X and Y play; then if X wins, Y sits down and the next game is between X and Z. Question: in the long run, which two players square off against each other most often? Least often? So far what I’ve described is completely realistic, but now we need to make a (strong) simplifying assumption. In practice people get tired and/or discouraged, so the probability that (say) X beats Y in any single round is probably not constant in time, but let’s pretend it is, to get a kind of baseline analysis: let 0 < pXY < 1 be the probability that X beats Y in any given game, and define 0 < pXZ < 1 and 0 < pY Z < 1 correspondingly. Consider the stochastic process P that keeps track of

Answers

Answer:

Step-by-step explanation:

(a) If the state space is taken as $S = \{(XY),(XZ),(YZ)\}$ , the probability of transitioning from one state, say (XY) to another state, say (XZ) will be the same as the probability of Y losing out to X, because if X and Y were playing and Y loses to X, then X and Z will play in the next match. This probability is constant with time, as mentioned in the question. Hence, the probabilities of moving from one state to another are constant over time. Hence, the Markov chain is time-homogeneous.

(b) The state transition matrix will be:

$P=\begin{vmatrix} 0 & p_(XY) & (1-p_(XY))\n p_(XZ)& 0& (1-p_(XZ))\n p_(YZ)&(1-p_(YZ)) & 0\end{vmatrix},$

where as stated in part (b) above, the rows of the matrix state the probability of transitioning from one of the states $S = \{(XY),(XZ),(YZ)\}$ (in that order) at time n and the columns of the matrix state the probability of transitioning to one of the states $S = \{(XY),(XZ),(YZ)\}$ (in the same order) at time n+1.

Consider the entries in the matrix. For example, if players X and Y are playing at time n (row 1), then X beats Y with probability $p_(XY)$ , then since Y is the loser, he sits out and X plays with Z (column 2) at the next time step. Hence, P(1, 2) = $p_(XY)$ . P(1, 1) = 0 because if X and Y are playing, one of them will be a loser and thus X and Y both together will not play at the next time step. $P(1, 3) = 1 - p_(XY)$ , because if X and Y are playing, and Y beats X, the probability of which is $1 - p_(XY)$ , then Y and Z play each other at the next time step. Similarly, $P(2, 1) = p_(XZ)$ , because if X and Z are playing and X beats Z with probability $p_(XZ)$ , then X plays Y at the next time step.

$vP = v,$

i.e., the steady state distribution v of the Markov Chain is such that after applying the transition probabilities (i.e., multiplying by the matrix P), we get back the same steady state distribution v. The Eigenvalues of the matrix P are found below:

$:det(P-\lambda I)=0\Rightarrow \begin{vmatrix} 0-\lambda & 0.6 & 0.4\n 0.975& 0-\lambda& 0.025\n 0.95& 0.05& 0-\lambda\end{vmatrix}=0$

$\Rightarrow -\lambda ^3+0.9663\lambda +0.0337=0\n\Rightarrow (\lambda -1)(\lambda ^2+\lambda +0.0337)=0$

The solutions are

$\lambda =1,-0.0349,-0.9651.$ These are the eigenvalues of P.

The sum of all the rows of the matrix $P-\lambda I$ is equal to 0 when $\lambda =1.$ Hence, one of the eigenvectors is :

$\overline{x} = \begin{bmatrix} 1\n 1\n 1 \end{bmatrix}.$

The other eigenvectors can be found using Gaussian elimination:

$\overline{x} = \begin{bmatrix} 1\n -0.9862\n -0.9333 \end{bmatrix}, \overline{x} = \begin{bmatrix} -0.0017\n -0.6666\n 1 \end{bmatrix}$

Hence, we can write:

$P = V * D * V^(-1)$ , where

$V = \begin{bmatrix} 1 & 1 & -0.0017\n 1 & -0.9862 & -0.6666 \n 1 & -0.9333 & 1 \end{bmatrix}$

and

$D = \begin{bmatrix} 1 & 0 & 0\n 0 & -0.9651 & 0 \n 0 & 0 & -0.0349 \end{bmatrix}$

After n time steps, the distribution of states is:

$v = v_0P^n\Rightarrow v = v_0(VDV^(-1))^n=v_0(VDV^(-1)VDV^(-1)...VDV^(-1))=v_0(VD^nV^(-1)).$

Let n be very large, say n = 1000 (steady state) and let v0 = [0.333 0.333 0.333] be the initial state. then,

$D^n \approx \begin{bmatrix} 1 & 0 & 0\n 0& 0 &0 \n 0 & 0 & 0 \end{bmatrix}.$

Hence,

$v=v_0(VD^nV^(-1))=v_0(V\begin{bmatrix} 1 & 0 & 0\n 0 &0 &0 \n 0& 0 & 0 \end{bmatrix}V^(-1))=[0.491, 0.305, 0.204].$

Now, it can be verified that

$vP = [0.491, 0.305,0.204]P=[0.491, 0.305,0.204].$

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Examine the work shown. Explain the error and find the correct result. 2(4 - 16) - (-30)2-12) - (-30) 24 - (-30) 54​

Answers

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Examine the work shown. Explain the error and find the correct result. 2(4 - 16) - (-30)2-12) - (-30)
24 - (-30)
54