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2 years ago

How many solutions does the linear system below have? -8x+2y=6 and -4x+y=3

Mathematics

1 answer:

Verizon [17]2 years ago

7 0

Answer:

Step-by-step explanation:

Try dividing the first equation by 3. The result will be identical to the second equation. Thus, we have two lines that coincide, and therefore there are an infinite number of solutions.

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People drive, on average, 14,579 miles per year. About how many mile each week is that? (Note: there are 52 weeks in a year)

Vikki [24]

Answer:278.44

Step-by-step explanation:

Divide the miles per year by months

6 0

2 years ago

Read 2 more answers

vincent’s copier can print in 7 hours. chris’s printer can do the same in 13 hours. how long would it take to do the job using b

Viefleur [7K]

Answer:

Step-by-step explanation:

1/7+1/13=1/t

multiply both sides by 7*13*t

13t+7t=7*13

21t=91

t=91/19

it will take 4 hours and 8 minutes

8 0

3 years ago

Please help me graph

Anika [276]

(x, y) = (-3, 4) hope this helps!

6 0

3 years ago

What is the answer to (z+1)(z+2)

stich3 [128]

Answer:

z^1+3z+2

Step-by-step explanation:

(z+1)(z+1)

Multiply each term in the first parenthesis by each term in the second parenthesis

Z x z+2z+z+2

Calculate the product

<u>z</u>^2 +2z+z+2

collect like terms

z^2+3z+2

2z+z

If a term doesnt have a coefficient it is considered that the coefficient is 1

2z+1z

(2+1)z

z^2+3z+2

4 0

3 years ago

Read 2 more answers

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

5 0

3 years ago