Answer:
element remained after 14 minutes = = 7.091 g =~ 10 g
Step-by-step explanation:
After every minute the amount remained will be
(100 - 26.9 ) % i.e. 73.1 %
which is 0.731 times as much as the amount was at the start.
if the number of minutes passed is represented by t the function f(t) represents the mass of element remaining our equation will be
f(t) = 570( 0.731) ^ t
t= 14 minutes
f(14) = 570 ( 0.731) ^ 14
= 7.091 g =~ 10 g
Answer:
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Step-by-step explanation:
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Answer:
152 lies between two consecutive integers 151 and 153.
Step-by-step explanation:
Let n be the given integer.
The predecessor of any integer n is given by n-1.
Similarly the successor of an integer n is given by n+1.
For example: Let n be 100
Then its predecessor is 100-1 = 99
Its successor is 100+1 = 101
In the given question, n=152.
Predecessor of 152 = 152 - 1 = 151
Successor of 152 = 152 + 1 = 153
So 152 lies between two successive integers 151 and 153.
So we know that
Year 1 = $55,000
Year 2 = 55000 + (55,000 x .04) = 55000 + 2200 = $57,200
Year 3 = 57200 + (57200 * .04) = 57200 + 2288 = $59,488
Year 4 = 59488 + (59488 * .04) = 59488 + 2379.52 = $61,867.52
Year 5 = 61867.52 + (61867.52 * .04) = 61867.52 + 2474.70 = $64,342.22
Year 6 = 64342.22 + (64342.22 * .04) = 64342.22 + 2573.69 = $66,915.91
Year 7 = 66915.91 + (66915.91 * .04) = 66915.91 + 2676.63 = $69,592.54
Year 8 = 69592.54 + (69592.54 * .04) = 69592.54 + 2783.70 = $72,376.24
Year 9 = 72376.24 + (72376.24 * .04) = 72376.24 + 2895.05 = $75,271.29
Year 10 = 75271.29 + (75271.29 * .04) = 75271.29 + 3010.85 = $78,282.14
So by the tenth year following graduation, the student would be making $78,282.14 a year. If you add up all of the values (Year 1 thru Year 10) you get a total income over ten years of $660,355.86