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3 years ago

6

Population of the world was 5.7 billion in 1995 and the observed growth rate was 2% per year. By what year will the population t

riple?

1 answer:

iren [92.7K]3 years ago

5 0

F(t) = a(1.02)t 11.4 = 5.7(1.02)t2 = 1.02tln 2 ≈ t ln 1.020.693 ≈ 0.0198 t35 = t

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A forest covers 72,000 acres. A survey finds that 0.7% of the forest is old-growth trees. How many acres of old-growth trees

Vsevolod [243]

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3 years ago

The length of a rectangle is 4 m more than the width the area of the rectangle is 45m^2 find the length and width

Novay_Z [31]

L=w+4m
A=w*l
A=w(w+4)=45
w^2+4w-45=0
w^2+9w-5w-45=0
w(w+9)-5(w+9)=0
(w+9)(w-5)=0

⇒ w+9=0 or w-5=0
⇒w-5=0
w=5
l=5+4
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4 years ago

In the following graph where is the function not increasing?

joja [24]

According to the given graph, the function is not increasing from x = -1 to x = 1.

Remember that not increasing behavior is expressed as a horizontal line.

<h2>Hence, the interval is [-1, 1].</h2> $-1\leq x\leq1$

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1 year ago

Twenty men can cut thirty trees in four hours. If four men leave the job, how many

Mice21 [21]

Answer:

As in 4 hours , 20 men can cut 30 trees. So, in 1 hour , 20 men can cut 30 trees / 4. So, in 6 hours, 20 men can cut 30 * 6 trees / 4 = 45 trees.

Step-by-step explanation:

done

8 0

3 years ago

A box contains eighteen $1 bills, ten $5 bills, eight $10 bills, three $20 bills, and one $100 bill. You blindly reach into the

zubka84 [21]

Answer: $7.70

Step-by-step explanation:

Given : A box contains eighteen $1 bills, ten $5 bills, eight $10 bills, three $20 bills, and one $100 bill.

Total bills = $18+10+8+3+1=40$

We know that probability of any event = $\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

i.e. the probability of getting $1 bills = $P(E_1)=\dfrac{18}{40}=0.45$

Probability of getting $10 bills = $P(E_2)=\dfrac{10}{40}=0.25$

Probability of getting $5 bills = $P(E_3)=\dfrac{8}{40}=0.2$

Probability of getting $20 bills = $P(E_4)=\dfrac{3}{40}=0.075$

Probability of getting $100 bills = $P(E_5)=\dfrac{1}{40}=0.025$

$\text{Expected value =}1\times P(E_1)+5\times P(E_2)+10\times P(E_3)+20\times P(E_4)+100\times P(E_5)\\\\=1\times 0.45+5\times 0.25+10\times0.2+20\times 0.075+100\times 0.025\\\\=7.7$

Hence, the expected value of your draw= $7.70

5 0

4 years ago