So since her balance must be at least 500, her ballance cannot reach 500
so 794-500=294
see how many 25's you can fit into 294
we know that 4 25's =100 and 294 is roughly 300
so 4 times 3=12
we minus one of the 25's because 294 is less than 300 so
the innequality is
>means more than
794-25x>500
$216 x 0.08 = $17.28.
Therefore $17.28 was collected for sales tax.
Answer:
a) 23.76%
b) 7.8%
Step-by-step explanation:
a) probability that a failure is due to loose keys.
loose key failure (27%) comes under mechanical failure(88%)
hence, probability that a failure is due to loose keys= 0.27×0.88= 0.2376= 23.76%
b) probability that a failure is due to improperly connected wire which comes under electrical failure = 0.12×0.13
probability that a failure is due to poorly welded wires which comes under electrical failure= 0.52×0.12
now, the probability that a failure is due to improperly connected or poorly welded wires. = 0.12(0.52+0.13)= 0.078= 7.8%
14C3 = 14! / 11!3!
<span>= 14 x13 x12 / 3x2x1 </span>
<span>= 2184 / 6 </span>
<span>= 364 different combinations </span>
<span>The first movie can be any of 14 </span>
<span>As you have already seen one the second movie can be any of the 13 remaining </span>
<span>As you have already seen two the third movie can be any of the 12 remaining. </span>
<span>Therefore there are 14 x 13 x 12 = 2184 PERMUTATIONS of movies you can see. </span>
<span>However among those 2184 different permutations will be instances where you have watched the same three movies but just in a different order. </span>
<span>eg ET, The Piano, Harry Potter = ET Harry Potter The Piano = The Piano, ET, Harry Potter = The Piano Harry Potter ET = Harry Potter ET The Piano = Harry Potter The Piano ET. </span>
<span>For each set of three films there are 3! or 3x2x1 or SIX different ways they can be arranged in. </span>
<span>Therefore we need to DIVIDE the above 2184 permutations by 6 to get the number of COMBINATIONS of different films that can be watched. </span>
<span>2184 / 6 = 364</span>
Let numbers of books be 'b' and numbers of CDs be 'c'
We can set up two equations:
Equation [1] ⇒

Equation [2] ⇒

We are solving for the number of books and the number of CDs bought
When we have two equations in terms of two different variables;

and

, that we need to solve, then this becomes a simultaneous equation problem.
First, rearrange Equation [1] to make either

or

the subject:


Then we substitute

into Equation [2]






Now we know the value of

which is

, substitute this value into

we have

Answer:
Numbers of books = 13
Numbers of CDs = 7