Have a shot it’s plainly obvious I know.

Assume the exponential growth model A(t)equals=Upper A 0 e Superscript ktA0ekt and a world population of 5.95.9 billion in 2006

Whitepunk [10]

Answer:

1.4% is the maximum acceptable annual rate of growth such that the population must stay below 24 billion during the next 100 years.

Step-by-step explanation:

We are given the following in the question:

The exponential growth model is given by:

$A(t) = A_0e^{kt}$

where k is the growth rate, t is time in years and $A_0$ is constant.

The world population is 5.9 billion in 2006.

Thus, t = 0 for 2006

$A_0 = 5.9\text{ billions}$

We have to find the maximum acceptable annual rate of growth such that the population must stay below 24 billion during the next 100 years.

Putting these values in the growth model, we have,

$24 = 5.9e^{100k}\\\\k = \dfrac{1}{100}\ln \bigg(\dfrac{24}{5.9}\bigg)\\\\k = 0.01403\\k = 0.01403\times 100\% = 1.4\%$

1.4% is the maximum acceptable annual rate of growth such that the population must stay below 24 billion during the next 100 years.

8 0

3 years ago

Multiply using a special product formula (6x-8) (6x+8)

ANEK [815]

The correct answer is B.

6x*6x=36x

-8*+8=-64

Thus, our equation is 36x-64

Hope this helps!

4 0

3 years ago

Please help me. math.

KiRa [710]

Answer:

y =293(1.06) ^x

y = 370 after 4 years

Step-by-step explanation:

If we are using the model for growth

y = a ( 1+b)^x

a is the initial population

b is the increase rate

We can substitute the values into the equation

y =293 (1+.06) ^ x

y =293(1.06) ^x

Let x equal 4 for the 4 years

y = 293(1.06)^4

y=369.9

3 0

3 years ago

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0

3 years ago

What is the range of the relation?

tia_tia [17]

The correct answer is (-2,1,2,8)

The range are the y- values

The explanation:

The range is the set of all second elements of ordered pairs (y-coordinates). Only the elements "used" by the relation or function constitute the range.

Hope this helped you :D

7 0

3 years ago