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3 years ago

StartFraction 47 Over 39 EndFraction = StartFraction b Over 2 EndFraction

Mathematics

1 answer:

statuscvo [17]3 years ago

3 0

Answer:

2.41

Step-by-step explanation:

Cross multiplication

47(2)=39b

94÷39=39b÷39

2.41=b

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A box contains 10 balls, of which 3 are red, 2 are yellow, and 5 are blue. Five balls are randomly selected with replacement. Ca

VLD [36.1K]

Answer:

The required probability is 0.1.

Step-by-step explanation:

red balls = 3

yellow balls = 2

blue balls = 5

Selected balls = 5

Number of elemnets in sample space = 10 C 5 = 1260

Ways to choose 1 red ball and 4 other colours = (3 C 1 ) x (7 C 4) = 105

Ways to choose 5 balls of other colours = 7 C 5 = 21

So, the probability is

$\frac{105}{1260} + \frac {21}{1260}\\\\\frac{126}{1260}=0.1$

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3 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$

The sum telescopes so that

$S_n=\dfrac2{14}-\dfrac2{5n+4}$

and as $n\to\infty$ , the second term vanishes and leaves us with

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17$

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