An account earns simple interest. (a) Find the interest earned. (b) Find the balance of the account.

1 answer:

5 0

1. The interest would be $332.80 and the balance of the account would be $2,932.80.

2. The interest would be $1,912.50 and the balance of the account would be $9,412.50

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A bag contains some white and black balls . The probability of picking two white balls one after other without replacement from

FromTheMoon [43]

Answer:

2/35

Step-by-step explanation:

Let w and b be the numbers of white and black balls in the bag respectively.

So, the total numbers of the balls in the bag is

$n=w+b\;\cdots(i)$

As the bag can hold maximum 15 balls only, so

$n\leq15 \;\cdots(ii)$

Probability of picking two white balls one after other without replacement

=Probability of the first ball to be white and the probability of second ball to be white

=(Probability of picking first white balls) x( Probability of picking 2nd white ball)

Here, the probability of picking the first white ball $=\frac{w}{n}$

After picking the first ball, the remaining

white ball in the bag $= w-1$

and the remaining total balls in the bag $=n-1$

So, the probability of picking the second white ball $=\frac{w-1}{n-1}$

Given that, the probability of picking two white balls one after other without replacement is 14/33.

$\Rightarrow \frac{w}{n} \times \frac{w-1}{n-1}=\frac{14}{33}$

$\Rightarrow \frac{w(w-1)}{n(n-1)} =\frac{14}{33}$

Here, $w$ and $n$ are counting numbers (integers) and 14 and 33 are co-primes.

Let, $\alpha$ be the common factor of the numbers $w(w-1)$ (numerator) and $n(n-1)$ (denominator), so

$\frac{w(w-1)}{n(n-1)} =\frac{14\alpha}{33\alpha}$

$\Rightarrow w(w-1)=14\alpha\cdots(iii)$

And $n(n-1)=33\alpha.$

As from eq. (ii), $n\leq 15$ , so, the possible value of $\alpha$ for which multiplication od two consecutive positive integers (n and n-1) is $33\alpha$ is 4.