Ask question

Ask question

All categories

1 year ago

10

The current population of a threatened animal species is 1.6 million, but it is declining with a half-life of 20 years. How many

animals will be left in 30 years? in 65 years?

1 answer:

GaryK [48]1 year ago

8 0

It is given that the half-life is 20 years and the current population is 1.6 million.

It is required to find the population in 30 and 65 years, respectively.

Recall the Exponential Decay Half-Life Formula:

$N=N_0\left(\frac{1}{2}\right)^{\frac{t}{h}}$

Where N₀ is the current population, t is the time in years, and h is the half-life.

(a) Substitute N₀=1.6, h=20, and t=30 into the formula:

$N=1.6\left(\frac{1}{2}\right)^{\frac{30}{20}}\approx0.6\text{ million}=600,000$

About 600,000 animals will be left in 30 years.

(b) Substitute N₀=1.6, h=20, and t=65 into the formula:

$N=1.6\left(\frac{1}{2}\right)^{\frac{30}{20}}\approx0.6\text{ million}=600,000$

You might be interested in

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

3 years ago

Ross drives 340 miles in 5 hour if ross continues to drive at the same rate how many miles can he drive in 10 hours?

yKpoI14uk [10]

He can drive about 680 miles in 10 hours.

7 0

3 years ago

A computer technician charges a one-time fee of $50 plus and additional $20 per hour of labor. If an equation is created to dete

allochka39001 [22]

In the equation the $50 would be the start up or in other words the y-intercept

4 0

3 years ago

OperumONS UNTUI a) Find the greatest number that divides 36, 45 and 63 without leaving a remainder. b) c) Find the greatest numb

larisa86 [58]

Answer:

a) The greatest number that divides 36, 45, and 63 without leaving a remainder is 9

b) The greatest number which exactly divides 90, 120, and 150 is 30

c) The greatest capacity of the bucket is 10 liters

d) The greatest number of people to whom the items can be distributed equally is 40 people

ii) 2 kg of wheat flour, 3 kg of corn and 4 kg of rice

e) The greatest number of children to whom the 48 orange, 80 bananas, and 144 apples can be distributed equally is 16

ii) 3 oranges, 5 bananas and 9 apples each

f) 5 mangoes at a time from the basket containing 120 mangoes

7 mangoes at a time from the basket containing 168 mangoes

g) The greatest length of each squared marble is 2 m

Step-by-step explanation:

a) The greatest number that divides 36, 45, and 63 without leaving a remainder = The highest common factor of 36, 45, and 63, which is given as follows;

36 = 9 × 4

45 = 9 × 5

63 = 9 × 7

Therefore, The greatest number that divides 36, 45, and 63 without leaving a remainder = The highest common factor of 36, 45, and 63 = 9

b) 90 = 30 × 3

120 = 30 × 4

150 = 30 × 5

The greatest number which exactly divides 90, 120, and 150 is 30

c) The factors of the volumes are;

50 l = 10 × 5 l

60 l = 10 × 6 l

70 l = 10 × 7 l

Therefore, the greatest capacity of the bucket = 10 liters

d) The masses of the items are

The factors of 80 = 40 × 2

120 = 40 × 3

160 = 40 × 4

Therefore the items can be distributed equally to 40 people

ii) Each person gets 2 kg of wheat flour, 3 kg of corn and 4 kg of rice

e) 48 = 16 × 3

80 = 16 × 5

144 = 16 × 9

Therefore, the greatest number of children = 16

ii) Each child gets 3 oranges, 5 bananas and 9 apples

f) The factors of 120 = 24 × 5

168 = 24 × 7

Therefore;

The greatest number of mangoes which is to be taken out of the basket with 120 mangoes = 5 mangoes each (24 times)

The greatest number of mangoes which is to be taken out of the basket with 168 mangoes = 7 mangoes each (24 times)

g) The area of the floor = 12 m × 10 m = 120 m²

The factor of 120 m² which is a perfect square is 4 therefore, we have;

The side length of each squared marble, s = √4 = 2

The side length of each squared marble, s = 2 m

7 0

3 years ago

Marrrta [24]

F(3) = t4(3) = 2

The value of the function at the point of expansion is the first (constant) term of the Taylor series.

6 0

4 years ago