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Anna71 [15]
3 years ago
8

A salesperson makes a commission of 1% of all cars that he sells. If his goal is to make at least $4,285 worth of commission dur

ing the month of January, how many dollars worth of cars does he need to sell? Write and solve an inequality to prove your answer.
Mathematics
1 answer:
shusha [124]3 years ago
6 0

Answer: $428,500

Step-by-step explanation:

Since commission is 1% = 1/100

Let the missing amount be represented as p

From the question, 1/100 × p = 4285

1/100p = 4285

Divide both sides by 1/100 to get the value of p

p = $4285 ÷ 1/100

p= $4285 × 100/1

p = $428500; i hope this helps, please mark as brainliest answer.

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The answer to the promblem is -7

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Which of the following is the product of the rational expression shown below? 5/2x-1 x 6/x
Rina8888 [55]

Answer:

30/(2x^2-x)

Step-by-step explanation:

5/(2x-1) * 6/x

Multiply the numerators

5*6 = 30

Multiply the denominatos

(2x-1) *x = 2x^2 -x


5/(2x-1) * 6/x = 30/x(2x-1) = 30/(2x^2-x)

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4x - 2y = 20<br> -8.x - 3y = 16
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for each problem, define the variable, write and inequality and find the solution. Ian needs to save at least $85.00 for a new p
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Answer:

he'll need to save for 5 weeks to get $85

Step-by-step explanation:

hope this helps :)

6 0
3 years ago
The computer center at Dong-A University has been experiencing computer down time. Let us assume that the trials of an associate
Schach [20]

Answer:

(a)0.16

(b)0.588

(c)[s_1$ s_2]=[0.75,$  0.25]

Step-by-step explanation:

The matrix below shows the transition probabilities of the state of the system.

\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)

(a)To determine the probability of the system being down or running after any k hours, we determine the kth state matrix P^k.

(a)

P^1=\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)

P^2=\begin{pmatrix}0.84&0.16\\ 0.48&0.52\end{pmatrix}

If the system is initially running, the probability of the system being down in the next hour of operation is the (a_{12})th$ entry of the P^2$ matrix.

The probability of the system being down in the next hour of operation = 0.16

(b)After two(periods) hours, the transition matrix is:

P^3=\begin{pmatrix}0.804&0.196\\ 0.588&0.412\end{pmatrix}

Therefore, the probability that a system initially in the down-state is running

is 0.588.

(c)The steady-state probability of a Markov Chain is a matrix S such that SP=S.

Since we have two states, S=[s_1$  s_2]

[s_1$  s_2]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[s_1$  s_2]

Using a calculator to raise matrix P to large numbers, we find that the value of P^k approaches [0.75 0.25]:

Furthermore,

[0.75$  0.25]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[0.75$  0.25]

The steady-state probabilities of the system being in the running state and in the down-state is therefore:

[s_1$ s_2]=[0.75$  0.25]

4 0
3 years ago
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