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gogolik [260]
3 years ago
14

Due to conservation efforts, the gray whale has been removed from the endangered species list. There are now 13,000 in the world

, and the population is expected to grow by 11% yearly. Write an equation that models this scenario.
Mathematics
1 answer:
brilliants [131]3 years ago
4 0

Answer:

P = 13,000(1.11) ^ t

Step-by-step explanation:

This scenario can be modeled using an exponential growth equation.

The exponential growth equations have the following form:

P = p (1 + r) ^ t

Where P is the population in year t

p is the initial population at t = 0

r is the growth rate

t is the time in years.

In this case we know that the current population is 13,000 and that the growth rate is 11%

So

p = 13,000\\r = 0.11

The equation that models this scenario is:

P = 13,000(1 + 0.11) ^ t

P = 13,000(1.11) ^ t

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Answer:

true

Step-by-step explanation:

8 0
3 years ago
What is the diameter and circumference of 6.7
irinina [24]

3.572

Area of a circle in terms of radius:

Area = π · r2 = 3.14 · 1.072 = 3.57 square meters.

In terms of diameter:

Area = π · (d

2

)2 = 3.14 · (2.13

2

)2 = 3.14 · (1.07)2 = 3.57 square meters.

In terms of circumference:

Area = C2

4π

= 6.72

4π

= 44.89

(4 · 3.14)

= 44.89

12.56

= 3.57 square meters.

Note: for simplicity, some results may be rounded to nearest hundredth and π was rounded to 3.14. See formula details below in this page.

A circle of radius = 1.066 or diameter = 2.133 or circunference = 6.7 meters has an area of 3.572 square meters which is equal to:

3.572E-6 square kilometers (km²)

35720 square centimeters (cm²)

0.0008827 acres (ac)

0.0003572 hectares (ha)

1.379E-6 square miles (mi²)

4.272 square yards (yd²)

38.45 square feet (ft²)

5537 square inches (in²)

Formulae:

Circle area formula

8 0
3 years ago
Read 2 more answers
12. The sales tax in Marvin's city is 7.33%. He bought a video game system for $299 and 2 games at $49 apiece. What was his tota
katen-ka-za [31]

Answer:

$426

Step-by-step explanation:

Step one:

given data

He bought a video game system for $299

2 games at $49 apiece= 2*49= $98

total cost = 299+98=$397

tax= 7.33%

=7.33/100*397

=0.0733*397

=29.1001

total cost = 29.1001+397

=$426

4 0
2 years ago
A biology class conducts a bird count every week during the semester. Using the number of species counted each​ week, a student
luda_lava [24]

Answer:

A student finds a​ 95% confidence interval of​ (16.34,18.69) for the mean number of species counted. This is a valid interval because the mean number of species or any population mean does not necessarily have to be a whole number, as stated by the student.

This given confidence interval of (16.34,18.69) helps us to simply estimate the mean species counted.

4 0
3 years ago
A fair 6-faced die is tossed twice. Letting E and F represent the outcomes of the each toss (which are independent), compute the
andreyandreev [35.5K]

Answer:

(a)  \frac{1}{18}            (d)  \frac{1}{9}

(b) \frac{1}{2}               (e) \frac{1}{3}

(c) \frac{4}{9}               (f)  \frac{1}{12}

Step-by-step explanation:

On tossing a 6-face die twice the outcomes of E and F are:

{1, 2, 3, 4, 5 and 6}

And there are total 36 outcomes of the form (E, F).

(a)

The sample space of getting a sum of 11 is: {(5, 6) and (6, 5)}

The probability  of getting a sum of 11 is:

P(Sum 11) =\frac{2}{36} \\=\frac{1}{18}

(b)

The sample space of getting an even sum is:

{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6) (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6) (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}

The probability  of getting an even sum is:

P(Even Sum)=\frac{18}{36}\\ =\frac{1}{2}

(c)

The sample space of getting an odd sum more than 3 is:

{(1, 4), (1, 6), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4),     (5, 6), (6, 1), (6, 3) and (6, 5)}

The probability of getting an odd sum more than 3 is:

<em>P</em> (Odd Sum more than 3) =

=\frac{16}{36}\\=\frac{4}{9}

(d)

The sample space of (E, F) such that E is even and less than 6, and F is odd and greater than 1 is:

{(2, 3), (2, 5), (4, 3) and (4, 5)}

The probability such that E is even and less than 6, and F is odd and greater than 1 is:

P(Even E1) =\frac{4}{36} \\=\frac{1}{9}

(e)

The sample space of E and F such that E is more than 2, and F is less than 4 is:

E = {3, 4, 5 and 6}     F = {1, 2 and 3}

Then the total outcomes of (E, F) will be 12.

The probability such that E is more than 2, and F is less than 4 is:

P(E>2, F

(f)

The sample space of (E, F) such that E is 4 and sum of E and F is odd is:

{(4, 1), (4, 3) and (4, 5)}

The probability such that E is 4 and sum of E and F is odd is:

P(E=4, E+F = Odd)=\frac{3}{36} \\=\frac{1}{12}

6 0
3 years ago
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