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Neko [114]
2 years ago
10

write an equation of the line that passes through the given point(-2,5) and is parallel to the given line 2y=4x-6

Mathematics
1 answer:
iragen [17]2 years ago
6 0
Y=2x+b

substitute -2 for x and 5 for y
5=2(-2)+b4

Solve for b
b=9

The equation is 
y=2x+9

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What multiples to -36 but adds to 16
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Read 2 more answers
Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand
dybincka [34]

Answer: Part a) P(a)=\frac{1}{\binom{r+w}{r}}

part b)P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = P(E)=\frac{Favourablecases}{TotalCases}

For part a)

Probability that a red ball is drawn in first attempt = P(E_{1})=\frac{r}{r+w}

Probability that a red ball is drawn in second attempt=P(E_{2})=\frac{r-1}{r+w-1}

Probability that a red ball is drawn in third attempt = P(E_{3})=\frac{r-2}{r+w-1}

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = P(E_{i})=\frac{r-i}{r+w-i}

Thus the probability that events E_{1},E_{2}....E_{i} occur in succession is

P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...

Thus P(E)=\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!

Thus our probability becomes

P(E)=\frac{1}{\binom{r+w}{r}}

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at (r+1)^{th} draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals \binom{r}{r-1}

Thus the probability becomes P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}

Thus required probability of case b becomes P(E)+ P(E_{i})

= P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\

7 0
3 years ago
What is the solution to 5/6x-1/3>1 1/3
CaHeK987 [17]

Answer:

Step-by-step explanation:

\frac{5}{6}x-1>1\frac{1}{3}\\\\\frac{5}{6}x-1>\frac{4}{3}\\\\\frac{5}{6}x>\frac{4}{3}+1\\\\\frac{5}{6}x>\frac{4}{3}+\frac{3}{3}\\\\\frac{5}{6}x>\frac{7}{3}\\\\x>\frac{7}{3}*\frac{6}{5}\\\\x>\frac{7*2}{5}\\x>\frac{14}{5}\\\\x>2\frac{4}{5}

3 0
3 years ago
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