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A single processor takes 2020 milliseconds (ms)(\text{ms}) to prepare data entries and 0.1n0.1n ms to copy the entries, where nn is the number of entries. A multiprocessor takes 7070 ms to prepare and copy one data entry, and whenever the number of entries is doubled the amount of time to prepare and copy them increases by 55 ms. Given 120120 ms to prepare and copy data entries, which processor type can prepare and copy more entries and how many more entries can it prepare and copy? \newlineChoose 11 answer:\newline(Choice A) The single processor can prepare and copy 176176 more entries.\newlineA The single processor can prepare and copy 176176 more entries.\newline(Choice B) The single processor can prepare and copy 989989 more entries.\newlineB The single processor can prepare and copy 989989 more entries.\newline(Choice C) The multiprocessor can prepare and copy 2424 more entries.\newlineC The multiprocessor can prepare and copy 2424 more entries.\newline(Choice D) The multiprocessor can prepare and copy 1,0121,012 more entries.\newlineD The multiprocessor can prepare and copy 1,0121,012 more entries.

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Posted 9 months ago

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The number of subscribers to a magazine is changing at a rate of r(t) r(t) subscribers per month (where t t is time in months).\newlineWhat does ∫810r′(t)dt=7 \int_{8}^{10} r^{\prime}(t) d t=7 mean?\newlineChoose 11 answer:\newline(A) As of month 1010 , the magazine had 77 subscribers.\newline(B) The average rate of change in subscribers between month 88 and month 1010 was 77 subscribers per month.\newline(C) The number of subscribers increased by 77 between t=8 t=8 and t=10 t=10 months.\newline(D) The rate of change of number of subscribers increased by 77 subscribers per month between t=8 t=8 and t=10 t=10 months.
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Posted 9 months ago

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A grocery store receives a shipment of bananas. The percent of the shipment that is still suitable for sale is decreasing at a rate of r(t) r(t) percent of the original shipment per day (where t t is the time in days). At t=2 t=2 , the grocery store found 85% 85 \% of the shipment to be suitable for sale.\newlineWhat does 0.85+∫24r(t)dt=0.6 mean?  0.85+\int_{2}^{4} r(t) d t=0.6 \text { mean? } \newlineChoose 11 answer:\newline(A) As of the fourth day, 60% 60 \% of the bananas suitable on the second day are still suitable.\newline(B) Between the second and fourth days, 60% 60 \% of the remaining bananas became unsuitable for sale per day.\newline(C) Between the second and fourth day, another 60% 60 \% of the bananas became unsuitable for sale.\newline(D) As of the fourth day, 60% 60 \% of the shipment was still suitable for sale.
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Posted 9 months ago

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Posted 9 months ago

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The population of a town grows at a rate of r(t) r(t) people per year (where t t is time in years). At t=3 t=3 , the town's population was 10001000 people.\newlineWhat does 1000+∫38r(t)dt=1500 1000+\int_{3}^{8} r(t) d t=1500 mean?\newlineChoose 11 answer:\newline(A) The town's population grew by 15001500 people between t=3 t=3 and t=8 t=8 .\newline(B) The town's population grew at a rate of 15001500 people per year at t=8 t=8 .\newline(C) The average rate at which the population grew between t=3 t=3 and t=8 t=8 is 15001500 people per year.\newline(D) At t=8 t=8 , the town's population was 15001500 people.
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Posted 9 months ago

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The value of a computer is decreasing at a rate that is proportional at any time to the value of the computer at that time.\newlineThe computer is worth $850 \$ 850 initially, and it is worth $306 \$ 306 after 22 years.\newlineHow much will the computer be worth after 55 years?\newlineChoose 11 answer:\newline(A) $0 \$ 0 \newline(B) $66 \$ 66 \newline(C) $79 \$ 79
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Posted 9 months ago

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Posted 9 months ago

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Carbon dating involves the measurement of the amount of carbon−14-14 in a sample. Archaeologists sometimes use carbon dating to estimate the number of years since an organism died. The function shows the ratio, R(t) R(t) , of carbon−14-14 remaining in a sample to the original amount of carbon−14-14 after t t years.\newlineR(t)=2−0.00175t R(t)=2^{-0.00175 t} \newlineIf one sample was 33 times as old as another sample of the same material, how would the carbon−14-14 ratios of the samples relate?\newlineChoose 11 answer:\newline(A) The ratio of the older sample would be 33 times the ratio of the newer sample.\newline(B) The ratio of the newer sample would be 33 times the ratio of the older sample.\newline(C) The ratio of the older sample would be the cube of the ratio of the newer sample.\newline(D) The ratio of the newer sample would be the cube of the ratio of the older sample.
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Posted 9 months ago

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D=1,874−0.55tD=1,874-0.55 t\newlineThe distance, DD, in meters, between an antarctic glacier and the coast tt days after January 11, 20102010 is approximated by the equation. How does the distance between the glacier and the coast change over time?\newlineChoose 11 answer:\newline(A) The glacier moves 0.550.55 meters per day closer to the shore.\newline(B) The glacier moves 1,8741,874 meters per day closer to the shore.\newline(C) The glacier moves 0.550.55 meters per day further from the shore.\newline(D) The glacier moves 1,8741,874 meters per day further from the shore.
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Posted 8 months ago

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Derin's boyfriend in Japan is $0.19\$0.19 per minute. The rate to call her family in Turkey is $0.12\$0.12 per minute. Derin wants to spend at least 3030 minutes calling her family in Turkey. Which of the following systems of inequalities best represents the relationship between JJ, the number of minutes Derin calls Japan, and TT, the number of minutes Derin calls Turkey? \newlineChoose 11 answer:\newline(A) 0.19J+0.12T≤250.19J+0.12T \leq 25\newlineT≤30T \leq 30\newline(B) 0.19J+0.12T≥250.19J+0.12T \geq 25\newlineT≤30T \leq 30\newline(C) 0.19J+0.12T≤250.19J+0.12T \leq 25\newlineT≥30T \geq 30\newline(D) 0.19J+0.12T≥250.19J+0.12T \geq 25\newlineT≥30T \geq 30
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Posted 8 months ago

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