Oore's law says that the number of transistors in a dense integrated circuit increases by 41%41\%41%41, percent every year. In 1974197419741974, a dense integrated circuit was produced with 5000500050005000 transistors.
Which expression gives the number of transistors in a dense integrated circuit in 1979197919791979?
Choose 1
Choose 1

(Choice A)
A
5000⋅0.4155000\cdot 0.41^55000⋅0.4155000, dot, 0, point, 41, start superscript, 5, end superscript
(Choice B)
B
5000+(1+0.41)55000+(1+0.41)^55000+( 1+0.41)55000, plus, left parenthesis, 1, plus, 0, point, 41, right parenthesis, start superscript, 5, end superscript
(Choice C)
C
5000(1+0.41)55000(1+0.41)^55000(1+0 .41)55000, left parenthesis, 1, plus, 0, point, 41, right parenthesis, start superscript, 5, end superscript
(Choice D)
D
5000+0.4155000+0.41^55000+0.4155000 , plus, 0, point, 41, start superscript, 5, end superscript

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!

answer: 180 degrees clockwise

step-by-step explanation: