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KengaRu [80]
3 years ago
5

How do you write seven hundred five thousandths

Mathematics
1 answer:
kotykmax [81]3 years ago
6 0
Seven hundred five thoudandths equal 0.705
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n 2000 the population of a small village was 2,400. With an annual growth rate of approximately 1.68%, compounded continuously,
Nitella [24]

Answer:

The population would be 3358

Step-by-step explanation:

Since, exponential growth function if the growth is compound continuously,

A = Pe^{rt}

Where,

P = initial population,

r = growth rate per period,

t = number of periods,

Given,

The population in 2000, P = 2,400,

Growth rate per year = 1.68% = 0.0168,

Number of years from 2000 to 2020, t = 20,

Thus, the population in 2020,

A=2400 e^{0.0168\times 20}

=2400 e^{0.336}

= 3358.4136

\approx 3358

6 0
3 years ago
PLEASE HELP!!!
DIA [1.3K]
You would need 27 more dollars because f(x)=7x+2 plug in the 5 in x’s spot and you get 37 and so subtract it from 10 to get 27
7 0
3 years ago
The equation V=13πr2h relates the volume V, the radius r, and the height h of a cone. Write h in terms of V and r.
maksim [4K]
V= \frac{1}{3} \pi r^2h \\  \\ 3V= \pi r^2h \\  \\ h= \cfrac{3V}{\pi r^2}
6 0
4 years ago
Read 2 more answers
Select all expressions that are equivalent to 64. 2 to the sixth power 2 to the eighth power 4 to the third power 8 squared 16 t
andreyandreev [35.5K]

Answer:

2 to the sixth, 4 to the third, 8 squared

Step-by-step explanation:  2*2*2*2*2*2=64 because you would have 2, 4,8,16,32,64

4*4*4=64 because you would have 4,16,64

8*8=64

3 0
3 years ago
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 10 Allen's hummingb
ruslelena [56]

Answer: provided in the explanation segment

Step-by-step explanation:

(a). from the question, we can see that since that б is known, we can use standard normal, z.

we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?

⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;

x ± z * б / √m

which is

3.15 ± 1.28 * 0.32/√10

= 3.15 ± 0.1295 = 3.0205 or 3.2795

(b). normal distribution of weight (c) б is known

(c). option (a) and (e) are correct

(d).  from the question, let sample size be given as S

this gives';

1.28 * 0.32/√S = 0.15

√S = (1.28 * 0.32) / 0.15 = 2.73

S = 7.4529

cheers i hope this helps

6 0
3 years ago
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