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4 years ago

(08.02)Which of the following graphs best represents the solution to the pair of equations below? y = x + 1

Mathematics

2 answers:

tankabanditka [31]4 years ago

7 0

The red line is represented by y = x + 1.

The blue line is represented by y = -x - 1.

The solution should be the coordinate (-1, 0).

lesantik [10]4 years ago

5 0

The graph of the two equations is "<span>A coordinate plane is shown with two lines graphed. </span>

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Solve: 62x - 3 = 6-2x+1

MatroZZZ [7]

Answer:

x = 5/32 = 0.156

Step-by-step explanation:

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4 years ago

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How do i solve this math stuff

Semenov [28]

A- y intercept= (0,5)
x intercept= (-5,0)
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3 years ago

Carter and his cousin are playing a game where they pick up colored sticks. Carter currently has 24 points

Fudgin [204]

The answer is 15 because I checked it with Socratic

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4 years ago

Determine if the problem below has a mistake or not. Identify the mistake and explain how to correct it.

pochemuha

Answer:

The problem has a mistake in it. The mistake is found in line 3. The person who solved the problem subtracted 6 from both sides when they should have added 6.

x = 1

Step-by-step explanation:

4x-2=8-6x

10x-2=8

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3 years ago

The computers of six faculty members in a certain department are to be re- placed. two of the faculty members have selected lapt

FinnZ [79.3K]

Answer:

A) The probability that both selected setups are for laptop computers is 0.067

B)The probability that both selected setups are desktop machines is 0.4

C)The probability that at least one selected setup is for a desktop computer is 0.933

D)The probability that at least on computer of each type is chosen for setup is 0.533

Step-by-step explanation:

Number of laptops = 2

Number of desktops = 4

Total number of outcomes = 15

a) what is the probability that both selected setups are for laptop computers?

Total number of outcomes = 15

So, the probability that both selected setups are for laptop computers = $\frac{1}{15}=0.067$

b)what is the probability that both selected setups are desktop machines?

Number of desktops = 4

Number of desktops to be chosen = 4

We will use combination

No. of ways to select two desktops = $^4C_2=\frac{4!}{2!(4-2)!}=6$

So,the probability that both selected setups are desktop machines= $\frac{6}{15}=0.4$

P(at least 1 desktop)=1-P(No desktop)

P(at least 1 desktop)=1-P(both laptops)

P(at least 1 desktop)=1-0.067=0.933

So,the probability that at least one selected setup is for a desktop computer is 0.933

d) what is the probability that at least on computer of each type is chosen for setup?

No. of ways to select one desktop $=^4C_1=\frac{4!}{1!(4-1)!}=4$

No. of ways to select one laptop = $^2C_1=\frac{2!}{1!(2-1)!}=2$

So, No. of ways to select one laptop and one desktop= $4 \times 2 = 8$

So,the probability that at least on computer of each type is chosen for setup= $\frac{8}{15}=0.533$

6 0

3 years ago