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9 days ago

12

From a bag containing 5 nickels, 8 dimes, and 7 quarters, 5 coins are drawn at random and all at once. What is the probability o

f getting 2 nickels, 2 dimes, and 1 quarter?

1 answer:

Svet_ta [4.3K]9 days ago

5 0

Answer:

0.13

Detailed explanation:

In a bag containing 5 nickels, 8 dimes, and 7 quarters, we randomly select 5 coins at once.

This means we need to pick 2 nickels from the 5 available, 2 dimes from the 8, and 1 quarter from the 7 quarters.

In total, there are 5 + 8 + 7 = 20 coins.

Selecting 5 coins from the entire set of 20 coins.

2 nickels can be chosen from 5 in 5C2 ways.

$5C2=\frac{5!}{2!(5-2)!} =\frac{5!}{2!(3!)!}=10$

2 dimes can be taken from 8.

$8C2=\frac{8!}{2!(8-2)!} =\frac{6!}{2!(6!)}=28$

1 quarter from 7 quarters.

$7C1=\frac{7!}{1!(7-1)!} =\frac{7!}{1!(6!)}=7$

5 coins chosen from 20.

$20C5=\frac{20!}{5!(15)!} =15504$

The probability of obtaining 2 nickels, 2 dimes, and 1 quarter is given by:

$\frac{10 \cdot 28 \cdot 7}{15504} =\frac{1960}{15504} =0.13$

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Which statement is accurate concerning the function’s discontinuities $f(x) = \frac{x-5}{3x^2-17x-28}$

A) There are gaps at x = 7 and.

B) Asymptotes exist at x = 7 and.

C) Asymptotes exist at x = –7 and.

D) Gaps are present at (–7, 0) and.

Answer:

B) Asymptotes exist at x = 7 and $(x = \frac{-4}{3})$

Step-by-step explanation:

Given:

$f(x) = \frac{x-5}{3x^2-17x-28}$

Goal:

Identify the correct statement

$f(x) = \frac{x-5}{3x^2-17x-28}$

We need to factor the denominator first.

$f(x) = \frac{x-5}{(3x+4)(x-7)}$

To express x in terms of (3x+4) and (x-7):

3x + 4 =

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Divide both sides by 3:

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Next, evaluate the limit when $(x = \frac{-4}{3})$ and at (x = 7)

lim f(x) as $(x = \frac{-4}{3})$ = ±∞

lim f(x) as (x=7) = ±∞

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It is established that

A correlation between two variables, x and y, demonstrates a direct variation if it can be written in the format $y/x=k$ or $y=kx$

For this question, we have

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$y/x=k$ -------> substitute ------> $k=12/5$

The equation is

$y=\frac{12}{5}x$

Keep in mind that

If a point is located on the direct variation line

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the point has to fulfill the direct variation equation

we will now validate each point

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Insert the values of x and y into the direct variation equation

$0=\frac{12}{5}*0$

$0=0$ -------> is valid

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the point $(0,0)$ lies on the direct variation line

case B) point $(2.5,6)$

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Insert the values of x and y into the direct variation equation

$6=\frac{12}{5}*2.5$

$6=6$ -------> is valid

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the point $(2.5,6)$ lies on the direct variation line

case C) point $(3,10)$

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Insert the values of x and y into the direct variation equation

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the point $(3,10)$ does not lie on the direct variation line

case D) point $(7.5,18)$

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Insert the values of x and y into the direct variation equation

$18=\frac{12}{5}*7.5$

$18=18$ -------> is valid

thus

the point $(7.5,18)$ lies on the direct variation line

case E) point $(12.5,24)$

$x=12.5\ y=24$

Insert the values of x and y into the direct variation equation

$24=\frac{12}{5}*12.5$

$18=30$ -------> is not valid

thus

the point $(12.5,24)$ does not lie on the direct variation line

case F) point $(15,36)$

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Insert the values of x and y into the direct variation equation

$36=\frac{12}{5}*15$

$36=36$ -------> is valid

thus

the point $(15,36)$ lies on the direct variation line

ultimately

the solution is

$(0,0)$

$(2.5,6)$

$(7.5,18)$

$(15,36)$

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