<span>Lets calculate an example:
Say, .001% of tires that come from the factory are bad. There is a 1/1000 chance that for any given tire randomly selected from the warehouse that a defect will be present. Each tire is a mutually exclusive independently occurring event in this case. The probability that a single tire will be good or bad, does not depend on how many tires are shipped in proportion to this known .001% (or 1/1000) defect rate.
To get the probability in a case like this, that all tires are good in a shipment of 100, with a factory defect rate of .001%, first divide 999/1000. We know that .999% of tires are good. Since 1/1000 is bad, 999/1000 are good. Now, multiply .999 x .999 x .999..etc until you account for every tire in the group of 100 shipped. (.999 to the hundredth power)
This gives us 0.90479214711 which rounds to about .90. or a 90% probability.
So for this example, in a shipment of 100 tires, with a .001% factory defect rate, the probability is about 90 percent that all tires will be good.
Remember, the tires are mutually exclusive and independent of each other when using something like a factory defect rate to calculate the probability that a shipment will be good.</span>
Step-by-step explanation:
Red=4
Blue=3
Yellow=3
Total=10
Pr(BY) without replacement= 3/10 *3/9
=6/90
=1/15
A) Solve 3x+6y=24 for x.
Subtracting 6y from both sides, we get
3x+6y-6y=24-6y
3x = -6y+24
Dividing both sides by 3, we get
x= -2y + 8
b) Solve 24−3x=6y for x.
Subtracting 24 from both sides, we get
24-24−3x=6y-24
-3x = 6y -24.
Dividing both sides by -3, we get
x= -2y + 8.
c) Solve 6y=24−3x for y in terms of x.
Dividing both side by 6.
y= 
d) Solve 24−6y=3x for x in terms of y.
Dividing both sides by 3, we get
8 - 2y = x.
x = -2y +8
<h3>Therefore, x=-2x+8 and
</h3>
Hmm
7
look at perfect squer befre and after
4=2^2
9=3^2
4<7<9
therefor
√4<√7<√9
2<√7<3
between 2 and 3
Answer:
Hello,
Step-by-step explanation:
heigth of the equilateral triangle:
Area of the triangle:
Area of the disk:
Probability:
