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IrinaK [193]
3 years ago
7

Which expression simplifies to 41? Choose all that apply!

Mathematics
1 answer:
Arada [10]3 years ago
5 0

Answer:

None of the expressions is simplified to 41.

Step-by-step explanation:

You have to follow the P.E.M.D.A.S., this is:

P:Parentheses first.

E:Exponent next.

M:Multiplication.

D:Division.

(Multiplication and Division is whichever come first, from left to right)

A: Addition.

S: Subtraction.

(Addition and Subtraction is whichever come first, from left to right).

<u><em>We are going to analyze every option:</em></u>

1. 10+ 2.4 - 1

First we have to solve the multiplication:

10+ 2.4 - 1=10+8-1

Now we have to solve the addition and subtraction from left to right:

10+8-1=18-1=17\neq 41

This is not a correct answer.

2. 10+2.(4-1)

We have to solve the parentheses first:

10+2.(4-1)=10+2.(3)

Now the multiplication and then the addition.

10+2.(3)=10+6=16\neq 41

This is not a correct answer.

3. (10+23).4-1

We have to solve the parentheses first:

(10+23).4-1=(33).4-1

Now the multiplication and then the subtraction:

(33).4-1=132-1=131\neq 41

This is not a correct option.

4. 10+(2^2.4)-1

We have to solve the parentheses first, in the parenthesis first we have to solve: 2^2 and then do the multiplication.

10+(2^2.4)-1=10+(4.4)-1=10+16-1

Now we have to solve the addition and subtraction from left to right:

10+16-1=26-1=25\neq 41

This expression is not a correct option.

Then none of the expressions is simplified to 41.

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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}}\\\\

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