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3 years ago

10 Points!!! Which comparison symbol makes each inequality statement true?

Mathematics

2 answers:

san4es73 [151]3 years ago

8 0

Less than
Greater than.
The two lines mean to only look at the number as it's distance from 0.

shutvik [7]3 years ago

7 0

-5.262<-5165

|-5.265| > |-5.165|

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How to find the area of a triangle with only the radius?

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Divide the bh by 1/2

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3 years ago

In a picture, there are 40 monkeys, and 60% of the monkeys are in the trees.

Zepler [3.9K]

The number of monkeys in the tree are 24.

<h3>What are percentages?</h3>

Percentage is the fraction of an amount expressed as a number out of hundred. The sign used to represent percentages is %.

<h3>How many monkey are in the tree?</h3>

Number of monkeys in the tree = percent x total number of monkeys

60% x 40

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2 years ago

Can someone who is good at math help? I only need 1a

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Answer:

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Step-by-step explanation:

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2 years ago

There are 8 pencils in a package.Two pencils from each pack will be given to the third grade.How many packages will be needed fo

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2 years ago

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