Answer:
a) There is a 66.7% chance that you were given box 1
b) There is a 80% chance that you were given box 1
Step-by-step explanation:
To find this, we need to note that there is a 1/10 chance of getting a defective bulb with box 1 and a 1/20 chance in box 2.
a) To find the answer to this, find the probability of getting a defective bulb for each box. Since there is only one bulb pulled in this example, we just use the base numbers given.
Box 1 = 1/10
Box 2 = 1/2
From this we can see that Box 1 is twice as likely that you get a defective bulb. As a result, the percentage chance would be 2/3 or 66.7%
b) For this answer, we need to square each of the probabilities in order to get the probability of getting a defective one twice.
Box 1 = 1/10^2 = 1/100
Box 2 = 1/20^2 = 1/400
As a result, Box 1 is four times more likely. This means that it would be a 4/5 chance and have a probability of 80%
This result is actually true for any exterior angle. The exterior angle of a triangle is equal to the sum of the two remote angles, and above is a short proof of it.
<span><span><span>3x</span>+<span>4y</span></span>=8
</span>Add -4y to both sides
<span><span><span><span>3x</span>+<span>4y</span></span>+<span>−<span>4y</span></span></span>=<span>8+<span>−<span>4y</span></span></span></span><span><span>3x</span>=<span><span>−<span>4y</span></span>+8
</span></span>Then you divide both sides by 3
<span><span><span>3x/</span>3</span>=<span><span><span>−<span>4y</span></span>+8/</span>3</span></span><span>x=<span><span><span><span>−4/</span>3</span>y</span>+<span>8/3
</span></span></span>And your answer is ...
<span>x=<span><span><span><span>−4/</span>3</span>y</span>+<span>8/<span>3</span></span></span></span>
A/B is a reduced fraction which can be represented as a
terminating decimal if and only b is of the form 2^n5^n where m and n are non
negative integers. For example: 7/250, the terminating decimal here is 0.028 as
250 as the denominator equals to 2*5^2.
According to the above, we must know whether the denominator
have only 2-s and or 5-s as the prime factors. So,
Q = 8 = 2 ^3 = hence the denominator has only 2 prime
factors. Fraction P/Q will be termed as terminated decimal which is sufficient.