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Yuri [45]
1 year ago
15

When rabbits were introduced to the continent of Australia they quickly multiplied and spread across the continent since there w

ere only primitive marsupial competitors and predators to interfere with the exponential growth of their population. The data in the following table can be used to create a model of rabbit population growth. Time (months)036912No. of Rabbits6321073097701. Find the regression equation for the rabbit population as a function of time x.2. Write the regression equation in terms of base e.3. Use the equation from part b to estimate the time for the rabbits to exceed 10,000.

Mathematics
1 answer:
stepan [7]1 year ago
3 0

Given:

We get the points (0,6), (3,32), and (12,770) from the given table.

1.

Let y be the number of the rabbit and x be the number of months.

Consider the regression equation of the form.

y=a(b)^x

Substitue x=0 and y=6 in the equation, we get

6=ab^0

a=6

Substitue x=12 and y=770 in the equation, we get

770=ab^{12}

\text{ Substitute }a=6\text{ in the equation, we get}

b^{12}=\frac{770}{6}

b^{}=\sqrt[3]{\frac{309}{32}}undefined

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seraphim [82]
-3
Because -2^2-8+1
-2to the second power is 4
4-8=-4
-4+1=-3
8 0
3 years ago
NEED HELP ASAP!!!<br> thank you.
Alex17521 [72]

Answer:

See explanation

Step-by-step explanation:

Consider triangles ACM and BCM. In these triangles,

  • m\angle 3=m\angle 4 - given;
  • m\angle 1=m\angle 2=90^{\circ} - definition of perpendicular lines CM and AB;
  • \overline{CM}\cong \overline{CM} - reflexive property.

So,

\triangle ACM\cong \triangle BCM by ASA postulate (if one side and two angles adjacent to this side of one triangle are congruent to one side and two angles adjacent to this side of another triangle, then two triangles are congruent).

Two-column proof:

      Statement                                 Reason

1. m\angle 3=m\angle 4                          Given

2. CM\perp AB                        Given

3. m\angle 1=m\angle 2=90^{\circ}       Definition of perpendicular lines CM and AB

4. \overline{CM}\cong \overline{CM}          Reflexive property

5. \triangle ACM\cong \triangle BCM                       ASA postulate

7 0
3 years ago
1 = 13 + 6n<br><br>could someone help??​
Mars2501 [29]
The answer is n = -2
8 0
3 years ago
Read 2 more answers
find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM​
Snezhnost [94]

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

  • Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>

  • To find - <u>Area </u><u>of </u><u>trapezium</u>

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

<u>Construction</u><u> </u><u>-</u>

draw \: CE \: \parallel \: AD \:  \\ and \: CD \: \perp \: AE

Now , we can clearly see that AECD is a parallelogram !

\therefore AE = CD = 13 cm

Now ,

AB = AE + BE \\\implies \: BE =AB -  AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm

Now , In ∆ BCE ,

semi \: perimeter \: (s) =  \frac{15 + 15 + 12}{2}  \\  \\ \implies \: s =  \frac{42}{2}  = 21 \: cm

Now , by Heron's formula

area \: of \: \triangle \: BCE =  \sqrt{s(s - a)(s - b)(s - c)}  \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)}  \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21}  \: cm {}^{2}

Also ,

area \: of \: \triangle \:  =  \frac{1}{2}  \times base \times height \\  \\\implies 18 \sqrt{21} =  \: \frac{1}{\cancel2}  \times \cancel12  \times height \\  \\ \implies \: 18 \sqrt{21}  = 6 \times height \\  \\ \implies \: height =  \frac{\cancel{18} \sqrt{21} }{ \cancel 6}  \\  \\ \implies \: height = 3 \sqrt{21}  \: cm {}^{2}

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

Area \: of \: trapezium =  \frac{1}{2}  \times(sum \: of \:parallel \: sides) \times height \\  \\ \implies \:  \frac{1}{2}  \times (25 + 13) \times 3 \sqrt{21}  \\  \\ \implies \:  \frac{1}{\cancel2}  \times \cancel{38 }\times 3 \sqrt{21}  \\  \\ \implies \: 19 \times 3 \sqrt{21}  \: cm {}^{2}  \\  \\ \implies \: 57 \sqrt{21}  \: cm {}^{2}

hope helpful :D

6 0
2 years ago
Solve for x. x+2/3=−1/6
Sphinxa [80]

Answer:-

x=-5/2

Step-by-step explanation:

x+2/3=-1/6

after cross-mutiplying,

6(x+2)=-3

6x+12=-3

6x=-3-12

6x=-15

x=-15/6

x=-5/2

hope this helps u!!

6 0
3 years ago
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