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3 years ago

10

Scott baked a batch of rolls. He gave a bag of 10 rolls to each of 7 friends he kept 1 roll for himself. How many rolls did he b

ake in all

2 answers:

USPshnik [31]3 years ago

7 0

10×7=70 70+1=71. 71 is the answer

Serjik [45]3 years ago

6 0

7 times 10 is 70, so in total the friends got that amount, but we aren’t done, because we still need to add the roll that he kept for himself, so it’s 71. Notice that it didn’t say he left a whole bag for himself, just one roll.

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Step-by-step explanation:

we know that

The equation of a exponential growth function is given by

$P(t)=a(1+r)^t$

where

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a is he initial value

r is the rate of change

Part 1) Australia

we have

$a=19,169,000\\r=0.6\%=0.6\100=0.006$

substitute

$P(t)=19,169,000(1+0.006)^t$

$P(t)=19,169,000(1.006)^t$

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Find the value of t

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substitute the value of t in the equation

$P(5)=19,169,000(1.006)^5=19,751,012\ people$

Part 2) China

we have

$a=1,261,832,000\\r=0.9\%=0.9\100=0.009$

substitute

$P(t)=1,261,832,000(1+0.009)^t$

$P(t)=1,261,832,000(1.009)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=1,261,832,000(1.009)^5=1,319,645,764\ people$

Part 3) Mexico

we have

$a=100,350,000\\r=1.8\%=1.8\100=0.018$

substitute

$P(t)=100,350,000(1+0.018)^t$

$P(t)=100,350,000(1.018)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=100,350,000(1.018)^5=109,712,539\ people$

Part 4) Zaire

we have

$a=51,965,000\\r=3.1\%=3.1\100=0.031$

substitute

$P(t)=51,965,000(1+0.031)^t$

$P(t)=51,965,000(1.031)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=51,965,000(1.031)^5=60,534,681\ people$

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