B) 64, because all these numbers have a difference of 14, and it you keep going, the 10th term will be 64.
<span>From the message you sent me:
when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths
If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function
Why does this work? Initially, you start with 500 mL of air that you breathe in, so
. After the second breath, you have 12% of the original air left in your lungs, or
. After the third breath, you have
, and so on.
You can find the amount of original air left in your lungs after
breaths by solving for
explicitly. This isn't too hard:
and so on. The pattern is such that you arrive at
and so the amount of air remaining after
breaths is
which is a very small number close to zero.</span>
We know that AB and CD are parallel. This allows many assumptions.
From that we know that angle A and angle D are congruent.
That means that x + 8 = 2x - 22 and we can solve for x
x + 8 = 2x - 22
x + 30 = 2x
30 = x or x = 30
We know from the figure that angle B is x or now that we solved for x is 30 degrees. Also, we know that both angle A and angle D are 38 degrees. Now we can solve for the vertical angle E which has a measure of y degrees. A triangle has the sum of its angles equal to 180 degrees.
We can set up an equation like this 30 + 38 + y = 180
30 + 38 + y = 180
68 + y = 180
y = 112 degrees
That is how you would solve this problem
Answer:
(4×-3)(2x-1)
Step-by-step explanation:
8x²-10x+3
8x²-4x-6x+3
4x(2x-1)-3(2x-1)
(4×-3)(2x-1)
Solution :
Mean time for an automobile to run a 5000 mile check and service = 1.4 hours
Standard deviation = 0.7 hours
Maximum average service time = 1.6 hours for one automobile
The z - score for 1.6 hours =
= 2.02
Now checking a normal curve table the percentage of z score over 2.02 is 0.0217
Therefore the overtime that will have to be worked on only 0.217 or 2.017% of all days.