Answer this plz,it’s due TODAY!
1 answer:
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<h3>Answer:</h3>
- <u>20</u> kg of 20%
- <u>80</u> kg of 60%
I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.
That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.
_____
<em>Using an equation</em>
If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...
... 0.60x + 0.20(100 -x) = 0.52·100
... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20
... x = 32/0.40 = 80 . . . . . kg of 60% alloy
... (100 -80) = 20 . . . . . . . .kg of 20% alloy
Answer:
a) 85.31%
b) 56 minutes
c) 17 days
d) 0.1721
Step-by-step explanation:
In order to make the calculations easier, let's <em>standardize the curve</em> by doing the change
where is the average trip-time and
the standard deviation and
P(X≤ t) is the probability that the trip takes less than t minutes.
a)
Here we are looking for the value of P(X>20).
For X=20 we have Z = (20-24)/3.8 = -1.05
So, we want the area under the standard normal curve for Z > -1.05
that we can compute either using a table or a computer and we find this area equals 0.8531
So, he arrives late to work 85.31% of the times.
(See picture 1 attached)
b)
In this case we are looking for a value t of time such that
P(X≥ t) = 20% = 0.2
So, we are seeking a value Z such that the area under the normal curve to the left of Z equals 0.2
By using a table or a computer we find Z = 0.842
(See picture 2 attached)
By inserting this value in the equation on standardization
So 20% of the longest trips takes 55.996 ≅ 56 minutes
c)
Since the average of days he arrives late is 85.31%, it is expected that in 20 work days he arrives late 85.31% of 20, which equals 17 days.
d)
Since the probability that he does not arrive late is 1-0.8531 = 0.1469 and the probability of arriving early is independent of the previous trip,
the probability of arriving early n days in a row is
0.1496*0.1496*...*0.1496 n times and
the probability of being early most than 10 trips in 20 days is
