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3 years ago

The lines (m-2)x+y=2 and mx+2y=k intersect at (2,8). Find the values of m and k

Mathematics

1 answer:

Lerok [7]3 years ago

4 0

Step-by-step explanation:

(m - 2)x + y = 2

y = -(m - 2)x + 2

8 = -(m - 2)(2) + 2

6 = -(m - 2)(2)

3 = -(m - 2)

-3 = m - 2

-1 = m

mx + 2y = k

2y = -mx + k

$y = - \frac{1}{2} mx + \frac{k}{2} \\ y = \frac{1}{2} x + \frac{k}{2} \\ 8 = \frac{1}{2} (2) + \frac{k}{2} \\ 7 = \frac{k}{2} \\ 14 = k$

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3 years ago

The slope of the line that passes through the points (-10,y) and (5, 2) is 2/5. What is the value of y?

Murrr4er [49]

Answer:

C, -4

Step-by-step explanation:

slope=change in y/change in x

$\frac{2 - y}{5 - ( - 10)} = \frac{2}{5}$

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2 years ago

If a full moon occurs on August 10, the next full moon would occur on/around _________.

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3 years ago

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The graph of a function increases twice as much over each subsequent equally sized interval. Which of the following could be the

Iteru [2.4K]

I hope this helps you

square root of (4x+41)=x+5

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3 years ago

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A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0

yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{10-x}$

Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

$P(x$

Where, for example, P(0) and P(1) are calculated as:

$P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098$

Then, $P(x$ , so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

$P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)$

Where, for example, P(4) is calculated as:

$P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001$

Then, $P(x\geq4 )=0.3504$ , so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

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3 years ago