Ask question

Ask question

All categories

2 years ago

10

In 1950, scientist estimated a certain animal population in a particular geographical area to be 6400 in 2000, the population ha

d risen to 7200. If this animal population experiences the same % increase over the next 50yrs what will be approximate population be?

1 answer:

irina1246 [14]2 years ago

7 0

Answer:

8,100

Step-by-step explanation:

7,200 - 6,400 = 800 increased in 50. Years

What percentage is 800 of 6,400 (%/100 = is/of)

x/100 = 800/6400 (cross multiply)

(800*100)/6400 = 12.50%

(12.50 * 7200)/100 = 900 increased

7,200 + 900 = 8,100

You might be interested in

Daud bought a plot of land with an area of 6 2/5 km. He divided the land equally into 8 small plots. What was the total area of

bezimeni [28]

Answer: $\dfrac{12}{5}\ km^2$

Step-by-step explanation:

Given

The area of land is $6\frac{2}{5}\ km^2$

Daud divide the land into 8 equal plots

So, the area of a single small plot is

$\Rightarrow \dfrac{\dfrac{30+2}{5}}{8}\\\\\Rightarrow \dfrac{32}{5\times 8}\\\\\Rightarrow \dfrac{4}{5}\ km^2$

The area of three such plots

$\Rightarrow 3\times \dfrac{4}{5}\\\\\Rightarrow \dfrac{12}{5}\ km^2$

3 0

3 years ago

What is the gcf of 99 and 54

Sergio [31]

Answer:

9

Step-by-step explanation:

The factors of 99 are 99, 33, 11, 9, 3, 1. The common factors of 54 and 99 are 9, 3, 1, intersecting the two sets above. In the intersection factors of 54 ∩ factors of 99 the greatest element is 9. Therefore, the greatest common factor of 54 and 99 is 9.

Hope this helps :)

Can I have brainliest?

6 0

3 years ago

Read 2 more answers

Laura paddles in her canoe at a speed of 5 mph for 5 miles. she then hops on her windsurfer and sails for 6 miles at a speed of

Serhud [2]

2 hours. What math is this?

6 0

3 years ago

What is the gcf of 48m^5n and 81m^2n^2

JulijaS [17]

Answer:

gcf is $3m^2n$

Step-by-step explanation:

we are given

$48m^5n$ and $81m^2n^2$

Firstly, we will find factors of both expression

$48m^5n=2\times 2\times2\times 2\times 3\times m^2\times m^3\times n$

$81m^2n^2=3\times 3\times 3\times 3\times m^2\times n\times n$

now, we will find common factors among them

we can see that

For finding gcf, we always find common factors

so,

common terms are 3,m^2 and n

so, gcf is

$=3\times m^2\times n$

$=3m^2n$

8 0

3 years ago

An SRS of 450 450 high school seniors gained an average of ¯ x = 20 x¯=20 points in their second attempt at the SAT Mathematics

maria [59]

Answer: (15.47, 24.53)

Step-by-step explanation:

We know that the confidence interval for population mean is given by :_

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

, where n= sample size.

$\sigma$ = standard deviation.

$\overline{x}$ = sample mean.

z*= Critical value.

Given : n= 450

$\overline{x}=20$

$\sigma=49$

Critical value for 95% confidence = z*=1.96 [From z-value table]

Then, the 95% confidence interval will be :-

$20\pm (1.96)\dfrac{49}{\sqrt{450}}$

$\approx 20\pm (1.96)(2.31)$

$\approx 20\pm 4.53$

$=(20-4.53,\ 20+4.53)=(15.47,\ 24.53)$

Hence, the 95% confidence interval for the mean change in score μ μ in the population of all high school seniors. : (15.47, 24.53)

3 0

3 years ago