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4 years ago

14

The graph shows one of the linear equations for a system of equations. Which equation represents the second linear equation for

the system of equations that has the solution which corresponds to a point at (- 16 5 , 14 5 )?

1 answer:

aliina [53]4 years ago

8 0

Answer: no the answer is 75

Step-by-step expl

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cooking instructions for a turkey say it should cook for 20 minutes per 500g, plus 15 minutes extra at the end. how long should

Dahasolnce [82]

Good evening ,

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Answer:

it needs 160 minutes =2hours and 40 minutes

___________________

Step-by-step explanation:

4kg=4000g

4000/500 = 8

Then the time needed is :

20×8= 160 minutes =2hours and 40 minutes.

:)

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4 years ago

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96/38n X 25/45 (multiplying rational expressions .simplify.)

Rzqust [24]

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$\frac{96}{38} \frac{25}{45}$

<h2>That's my answer I hope it helps #brainliestbunch - _ - ;) </h2>

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3 years ago

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WILL GIVE BRANLIEST! FINAL QUESTION

blondinia [14]

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8 0

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

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

4 years ago

I have this same problem in math but with different wording. I thought it would be funnier to show this one though xD but pls he

Sever21 [200]

Wish I could help. If you give me the real problem I will help you.

5 0

4 years ago