Initial population of rainbow smelt: 227.
Initial population of bloater fish: 1,052.
t - number of the years;
The equation is:
227 - 19.76 t = 1,052 - 92.57 t
92.57 t - 19.76 t = 1,052 - 227
72.81 t = 825
t = 825 : 72.81
t = 11.33 years = 11 1/3 years = 11 years 4 months
Answer: After 11 years and 4 months.

5 0

3 years ago

Read 2 more answers

3. What is the net amount that a contestant should expect to win per game if the game were to be played many

larisa [96]

Answer:

0.55

Missing Statement:

At a carnival, one game costs $1 to play. The contestant gets one shot in an attempt to bust a balloon. Each balloon

contains a slip of paper with one of the following messages:

Sorry, you do not win, but you get your dollar back. (The contestant has not lost the $1 cost.)
Congratulations, you win $2. (The contestant has won $1.)
Congratulations, you win $5. (The contestant has won $4.)
Congratulations, you win $10. (The contestant has won $9.)

If the contestant does not bust a balloon, then the $1 cost is forfeited. The table below displays the probability

distribution of the discrete random variable, or net winnings for this game

Net Winnings −1 0 1 4 9

<u><em>Probability</em></u> 0.25 ? 0.3 0.08 0.02

Step-by-step explanation:

First we find of the missing probability for net winnings of 0, in which case:

When you receive the message, “Sorry, you do not win, but you get your dollar back,” then your net winnings is 0$.

The probability of winning 0$ is = 1 - 0.25 - 0.3 - 0.08 -0.02 = 0.35

So now we have the complete details:

Net Winnings −1 0 1 4 9

Probability 0.25 0.35 0.3 0.08 0.02

Now we need to find ,the net amount that a contestant should expect to win is the expected value of the probability distribution.

Expected value of probability distribution = −1(0.25 ) +0 (0.35 ) + 1(0.3 ) +4 (0.08 ) + 9(0.02 )

Expected value of probability distribution = 0.55

<u><em>The net amount that a contestant should expect to win per game if the game were to be played many times is 0.55</em></u>

7 0

4 years ago

Suppose u = (1,2,3), v = (2, -3, -4), and w = (1,0,-2). (a) Find u-2v + 3w. (b) Write the vector (3,4,5) as a linear combination

marishachu [46]

<h2>Answer with explanation:</h2>

We are given vectors u,v and w as follows:

$u=(1,2,3)\,\ v=(2,-3,-4),\ and\ w=(1,0,-2)$

(a)

Now we are asked to find the value of:

$u-2v+3w$

The value is calculated as follows:

$u-2v+3w=(1,2,3)-2(2,-3,-4)+3(1,0,-2)\\\\i.e.\\\\u-2v+3w=(1,2,3)+(-2\times 2,-2\times -3,-2\times -4)+(3\times 1,3\times 0,3\times -2)\\\\i.e.\\\\u-2v+3w=(1,2,3)+(-4,6,8)+(3,0,-6)\\\\i.e.\\\\u-2v+3w=(1-4+3,2+6+0,3+8-6)\\\\i.e.\\\\u-2v+3w=(0,8,5)$

Hence,

$u-2v+3w=(0,8,5)$

(b)

We are asked to represent (3,4,5) as a linear combination of u, v, and w.

Let:

$(3,4,5)=au+bv+cw$

$(3,4,5)=a(1,2,3)+b(2,-3,-4)+c(1,0,-2)$

i.e.

$(3,4,5)=(a,2a,3a)+(2b,-3b,-4b)+(c,0,-2c)\\\\i.e.\\\\(3,4,5)=(a+2b+c,2a-3b,3a-4b-2c)\\\\i.e.\\\\a+2b+c=3----------(1)$

$2a-3b=4-------------(2)$

$3a-4b-2c=5------------(3)$

on multiplying equation(1) by 2 an then adding equation (1) to equation (3)

$5a=11\\\\i.e.\\\\a=\dfrac{11}{5}$

on putting the value of a in equation (2) we get:

$2\times \dfrac{11}{5}-3b=4\\\\i.e.\\\\\dfrac{22}{5}-3b=4\\\\i.e.\\\\3b=\dfrac{22}{5}-4\\\\i.e.\\\\3b=\dfrac{2}{5}\\\\i.e.\\\\b=\dfrac{2}{15}$

Now on putting the value of a and b in equation (1)

$\dfrac{11}{5}+2\times \dfrac{2}{15}+c=3\\\\i.e.\\\\\dfrac{11}{5}+\dfrac{4}{15}+c=3\\\\i.e.\\\\c=3-\dfrac{11}{5}-\dfrac{4}{15}\\\\i.e.\\\\c=\dfrac{3\times 15-11\times 3-4}{15}\\\\i.e.\\\\c=\dfrac{45-33-4}{15}\\\\i.e.\\\\c=\dfrac{8}{15}$

Hence, we have:

$(3,4,5)=\dfrac{11}{5}(1,2,3)+\dfrac{2}{15}(2,-3,-4)+\dfrac{8}{15}(1,0,-2)\\\\i.e.\\\\(3,4,5)=\dfrac{11}{5}u+\dfrac{2}{15}v+\dfrac{8}{15}w$

5 0

4 years ago

Solve for the missing angle

nata0808 [166]

Answer:

15.9778

≈ 16

Step-by-step explanation:

Formula= a² + b² = c²

7.7² + 14² = 255.29

√255.29 = 15.9778 or ≈16

8 0

3 years ago