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2 years ago

Need help with 2&3 please .

Mathematics

1 answer:

Vsevolod [243]2 years ago

5 0

Answer:

brainly.com/question/12500028

Step-by-step explanation:

brainly.com/question/12500028

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38 ×29 use rounding to estimate each product

nydimaria [60]

38 rounds to 40.
29 rounds to 30.
40 x 30 = 120.

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2 years ago

Express 110111 base two to base ten

lidiya [134]

Answer:

Step-by-step explanation:

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2 years ago

Express the following pattern as an algebraic expression: 88, 75, 62, 49. Use “n” for the term number.

Ostrovityanka [42]

Answer:

-13n + 101

Step-by-step explanation:

88, 75, 62, 49. = -13 -13 -13 = -13n + 101 (as 88+13 = 101)

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1 year ago

Read 2 more answers

The perimeter of a rectangle is given by P equals 2 l plus 2 w. Find the width of a rectangle whose length is 35 and perimeter i

maks197457 [2]

2l + 2w = P
l = 35, P = 84, w=?

2*35 + 2*w = 84
70 + 2w = 84
2w = 84 - 70
2w = 14
w = 14/2
w = 7 in

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

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

2 years ago