Which expression is equivalent to −16y+13 ?
2 answers:
You might be interested in
Answer:
x = -4, 5/2
Step-by-step explanation:
A quadratic can be solved may ways, including graphing, factoring, and the quadratic formula. You can also check possible answers by making use of the relationships between solutions and the coefficients.
__
A graph is attached. It shows the solutions to be -4 and 5/2.
__
When factored, the equation becomes ...
(2x -5)(x +4) = 0 . . . . . has solutions x=-4, x=5/2 (these make the factors zero)
__
Using the quadratic formula, the solutions of ax^2 +bx +c = 0 are found from ...
x = (-b±√(b²-4ac))/(2a)
x = (-3±√(3²-4(2)(-20))/(2(2)) = (-3±√169)/4 = {-16, +10}/4
x = {-4, 5/2}
__
For ax^2 +bx +c = 0, the solutions must satisfy ...
product of solutions is c/a = -20/2 = -10
Only the first and last choices have this product.
sum of solutions is -b/a = -3/2
Only the first choice (-4, 5/2) has this sum.
Answer:
a) 59.34%
b) 44.82%
c) 26.37%
d) 4.19%
Step-by-step explanation:
(a)
There are in total <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>
As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.
So, the probability of selecting exactly 2 bulbs of 75 W is
(b)
The probability of selecting three 40-W bulbs is
The probability of selecting three 60-W bulbs is
The probability of selecting three 75-W bulbs is
Since <em>the events are disjoint</em>, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%
(c)
There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is
(d)
The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, <em>supposing there is no replacement</em>, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.
As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is
Since there are no replacement, the probability of taking a second non 75-W bulb is now
Following this procedure 5 times, we find the probabilities
which are
0.6, 0.5714, 0.5384, 0.5, 0.4545
As the events are independent, the probability of choosing 5 non 75-W bulbs is the product
0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%