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3 years ago

Help will be greatly appreciated!!! Write the system of equations for these two lines.

Mathematics

1 answer:

Darya [45]3 years ago

5 0

Answer:

See answer below:

Step-by-step explanation:

Blue: y=2x+1

Red: y=3x-2

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What is 6+9+4+1+10? Please explain?

Viktor [21]

30, 6+9= 15+4= 19+1= 20+10= 30

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3 years ago

Read 2 more answers

Nina had 17 marbles in a bag. Her mother put a handful of marbles in the bag. When Nina counts her marbles, she discovers she no

Ksivusya [100]

The second equation is correct. Nina initally had 17 marbles in her bag. Her mother placed m marbles into her bag. After counting, she now has 34 marbles. Part of this 34 marbles is already her initial 17 marbles. Therefore, 17 marbles initially + m unknown marbles = 34 marbles. The equation <span>17 + m = 34 is the answer.</span>

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3 years ago

Help help help plz plz plz

DaniilM [7]

Answer:

Yes

Step-by-step explanation:

What would you do if he said ok so he said yes would go?

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3 years ago

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 $\binom{r}{r-1}$