All categories

2 years ago

There are 200 birds in a flock. 120 of them are female. what percentage of flock are male

Mathematics

1 answer:

drek231 [11]2 years ago

4 0

200 birds - 120 females = 80 males

80 males ÷ 200 birds = 0.4

0.4 × 100 = 40%

You might be interested in

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago

Please solve (-3×+15)+(-3×+2)

koban [17]

(-3x + 15) + (-3x + 2)
Simplify by combining like terms.
-6x + 17

-6x + 17

3 0

2 years ago

Read 2 more answers

An adult meerkat weighed 0.776 kilograms . this is 7.47 hectograms more than the weight of a baby meerkat . how much dose a baby

Firlakuza [10]

Let x = weight of adult meerkat, y=weight of baby meerkat, Find y.
x=0.776 kg
y+7.47hectogram=x=0.776
Since 1 hectogram = 0.1 kg,
Therefore the weight of baby meerkat is, y= 0.029 kg

7 0

2 years ago

Read 2 more answers

Darryl deposits $1,500 into a savings account that has a simple interest rate of 2.7%. Lori deposits $1,400 into a savings accou

igor_vitrenko [27]

Darryl:
Answer:
A = $1,905.00

(I = A - P = $405.00)

Equation:
A = P(1 + rt)

Lori:

Answer:
A = $1,932.00

(I = A - P = $532.00)

Equation:
A = P(1 + rt)

Thus $532-$405= $127 more in Lori's account

3 0

2 years ago

Read 2 more answers

Three years ago Tom was twice as old as Jean, and in two years the sum of their ages will be 28 years. Find their present ages.

lubasha [3.4K]

Answer:

Jean is 9 and Tom is 15.

Step-by-step explanation:

3 years ago, Tom was 12 and Jean was 6, hence Tom was twice as old as Jean.

Since that was their age 3 years ago, they are currently 15 (Tom) and 9 (Jean).

Add 2 years to each of these ages, you get 17 and 11.

17 + 11 = 28