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Paladinen [302]

3 years ago

9

Allison plays basketball and last season she made a total of 87 shots (two- or three- point shots). Her points total was 191. Ho

w many three-points shots did she make

1 answer:

yarga [219]3 years ago

4 0

Answer:

17

Step-by-step explanation:

Let the number of 2 point shots be t.

Let the number of 3 point shots be h.

Allison made 87 shots and scored 191 total points. This implies two things:

t + h = 87 ________(1)

2t + 3h = 191 ______(2)

Let us eliminate t by multiplying (1) by 2 and subtracting from (2):

=> 2t + 3h = 191

- <u>2t + 2h = 174</u>

<u> h = 17 </u>

<u />

Therefore, she made 17 three point shots.

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3 years ago

A town has 5000 people in year t = 0. Calculate how long it takes for the population P to double once, twice, and three times, a

UNO [17]

Answer:

Step-by-step explanation:

At the time t = 0, population of the town = 5000

Rate of population increase = 500 per year

Therefore, the equation that will represent the population will be

$P_{t}=P_{0}+500t$

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By plugging in the values in the equation,

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Population will be = 10000×2 = 20000

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Population of the town = 20000×2 = 40000

Now we plug in the values in the equation,

40000 = 5000 + 500t

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t = $\frac{35000}{500}$

t = 70 years

b). If the population growth is 5%.

Then the growth will be exponential represented by

$T_{n}=T_{0}(1+\frac{r}{100})^{t}$

$T_{n}$ = Population after t years

$T_{0}$ = Initial population

t = time in years

For double once,

Population after t years = 10000

$10000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{10000}{5000}$

$(1.05)^{t}=2$

Take log on both the sides

$log(1.05)^{t}=log2$

tlog(1.05) = log2

t = $\frac{log2}{log1.05}$

t = 14.20 years

For double twice,

Population after t years = 20000

$20000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{20000}{5000}$

$(1.05)^{t}=4$

Take log on both the sides

$log(1.05)^{t}=log4$

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t = $\frac{log4}{log1.05}$

t = 28.413 years

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Population after t years = 40000

$40000=5000(1+\frac{5}{100})^{t}$

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Take log on both the sides

$log(1.05)^{t}=log8$

tlog(1.05) = log8

t = $\frac{log8}{log1.05}$

t = 42.620 years

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Answer:

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