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3 years ago

2.

Mathematics

1 answer:

abruzzese [7]3 years ago

5 0

Answer:

a) 30 kangaroos in 2030

b) decreasing 8% per year

c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo

Step-by-step explanation:

We assume your equation is supposed to be ...

P(t) = 76(0.92^t)

a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30

In the year 2030, the population of kangaroos in the province is modeled to be 30.

b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.

The population is decreasing by 8% each year.

c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.

P(100) = 75(0.92^100) = 76(0.0002392)

P(100) ≈ 0.0182, about 1/55th of a kangaroo

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An electrician charges a $100 flat fee, plus $75 per hour to work on a house.

jek_recluse [69]

Answer:

y=75x+100

Step-by-step explanation:

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2 years ago

A BAG CONTAINS 10 RED BALLS AND FIVE BLUE BALLS WHAT IS TGE CHANCE OF PICKING A BLUE BALL.PLEASE EXPLAIN HOW TO SOLVE IT

Nimfa-mama [501]

Answer:

5/15

Step-by-step explanation:

there are 15 balls altogether, 5 of which are blue meaning there is a 5/15 chance of picking a blue ball which as a percentage is 33.33% chance

4 0

3 years ago

The sum of the first three terms in a GP is 38.Their product is 1728.Find the values of the three terms.

dusya [7]

Answer:

The value of the three terms is 8 and 18

Step-by-step explanation:

Let "a" be the first term and "r" be the common ratio.

Then from the condition, we have these two equations

a + ar + ar^2 = 38, (1)

a*(ar*)*(ar^2) = 1728. (2)

From equation (2), a^3*r^3 = 1728, or (ar)^3 = 1728, which implies

ar = root%283%2C1728%29 = 12; (3)

hence,

r = 12%2Fa. (4)

Now, in equation (1) replace the term ar by 12, based on (3). You will get

a + 12 + ar^2 = 38, which implies

a + ar^2 = 26. (5)

Next, substitute r = 12%2Fa into equation (5), replacing "r" there. You will get

a + a%2A%28144%2Fa%5E2%29 = 26, or

a + 144%2Fa = 26.

Multiply by "a" both sides and simplify

a^2 - 26a + 144 = 0,

%28a-13%29%5E2 - 169 + 144 = 0

%28a-13%29%5E2 = 25

a - 13 = +/- sqrt%2825%29 = +/- 5.

Thus two solutions for "a" are a = 13 + 5 = 18 or a = 13 - 5 = 8.

If a = 8, then from (4) r = 12%2F8 = 3%2F2.

If a = 18, then from (4) r = 12%2F18 = 2%2F3.

In the first case, if a = 8, then the three terms are 8, 8%2A%283%2F2%29 = 12 and 8%2A%283%2F2%29%5E2 = 18.

In this case, the sum of terms is 8 + 12 + 18 = 38, so this solution does work.

In the second case, if a = 18, then the three terms are 18, 18%2A%282%2F3%29 = 12 and 18%2A%282%2F3%29%5E2 = 8.

In this case, the sum of terms is 18 + 12 + 8 = 38, so this solution does work, too.

ANSWER. The problem has two solution:

a) first term is 18; the common difference is 2%2F3 and the progression is 18, 12, 8.

b) first term is 8; the common difference is 3%2F2 and the progression is 8, 12, 18.

4 0

2 years ago

What is 1 divided by 998,001 to 9 decimal places?

Otrada [13]

Trick question? :P

1/998,001 should be:

0.000001002 or

1.00200300*10^-6

7 0

3 years ago

The vertices A, B, C of a triangle are (2,1),(5,2),and (3,4) respectively . Find the coordinate of the circumcentre and also the

BlackZzzverrR [31]

Answer:

Circumcenter = (4,0)

Circumcircle = √5

Step-by-step explanation:

The circumcentre is the point of intersection of the perpendicular bisectors of a triangle. The vertices of the triangle are equidistant to the circumcentre.

Let us assume the coordinate of the circumcentre is at O(x, y). Therefore the distance between the cirmcumcenter and the vertices are:

$AO=\sqrt{(x-2)^2+(y-1)^2} =\sqrt{x^2-4x+4+(y^2-2y+1)}\\=\sqrt{x^2+y^2-4x-2y+5} \\\\BO=\sqrt{(x-5)^2+(y-2)^2} =\sqrt{x^2-10x+25+(y^2-4y+4)}\\=\sqrt{x^2+y^2-10x-4y+29} \\\\CO=\sqrt{(x-3)^2+(y-4)^2} \\=\sqrt{x^2-9x+9+(y^2-8y+16)}=\sqrt{x^2+y^2-9x-8y+25} \\$

AO = BO, therefore

√(x² + y²-4x-2y+5) = √(x² + y² - 10x - 4y + 29)

x² + y²-4x-2y+5 = x² + y² - 10x - 4y + 29

6x + 2y = 24 (1)

BO = CO

√(x² + y² - 10x - 4y + 29) = √(x² + y² - 9x - 8y + 25)

x² + y² - 10x - 4y + 29 = x² + y² - 9x - 8y + 25

-x + 4y = -4 (2)

Multiply equation 2 by 6 and add to equation 1:

26y = 0

y=0

Put y = 0 in -x + 4y = -4

-x + 4(0) = -4

x = 4

The cicumcenter is at (4, 0)

The radius of the circumcircle = AO = BO = CO. Therefore:

$Radius=AO=\sqrt{(4-2)^2+(0-1)^2} =\sqrt{4+1}=\sqrt{5}$

7 0

3 years ago