1. Name of shape<br> Volume formula<br> u! 8<br> 12 in<br> Volume

The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) =

Dafna1 [17]

Answer:

a) L'(t) = 34.416*e^(-0.18*t)

b) L'(0) = 34 cm/yr , L'(1) =29 cm/yr , L'(6) =12 cm/yr

c) t = 10 year

Step-by-step explanation:

Given:

- The length of fish grows with time. It is modeled by the relation:

L(t) = 200*(1-0.956*e^(-0.18*t))

Where,

L: Is length in centimeter of a fist

t: Is the age of the fish in years.

Find:

(a) Find the rate of change of the length as a function of time

(b) In this part, give you answer to the nearest unit. At what rate is the fish growing at age: t = 0 , t = 1, t = 6

c) When will the fish be growing at a rate of 6 cm/yr? (nearest unit)

Solution:

- The rate of change of length of a fish as it ages each year can be evaluated by taking a derivative of the Length L(t) function with respect to x. As follows:

dL(t)/dt = d(200*(1-0.956*e^(-0.18*t))) / dt

dL(t)/dt = 34.416*e^(-0.18*t)

- Then use the above relation to compute:

L'(t) = 34.416*e^(-0.18*t)

L'(0) = 34.416*e^(-0.18*0) = 34 cm/yr

L'(1) = 34.416*e^(-0.18*1) = 29 cm/yr

L'(6) = 34.416*e^(-0.18*6) = 12 cm/yr

- Next, again use the derived L'(t) to determine the year when fish is growing at a rate of 6 cm/yr:

6 cm/yr = 34.416*e^(-0.18*t)

e^(0.18*t) = 34.416 / 6

0.18*t = Ln(34.416/6)

t = Ln(34.416/6) / 0.18

t = 10 year

7 0

3 years ago

public accountant financial planner 50.2 48.0 59.8 49.2 57.3 53.1 59.2 55.9 53.2 51.9 56.0 53.6 49.9 49.7 58.5 53.9 56.0 52.8 51

ddd [48]

Using the formula of test statistic, the value of the test statistic is 2.6961.

The mean of the public accountant is

Mean x(p) =50.2+59.8+57.3+59.2+53.2+56.0+49.9+58.5+56.0+51.9/10

x(p) = 55.2

Now the standard deviation of public accountant is

SD(p) = √{∑(x-x(p))^2/n-1}

SD(p) = √(50.2-55.2)^2+(59.8-55.2)^2+..................+(51.9-55.2)^2/n-1

After solving;

SD(p) = 3.34

The mean of the financial planner is

Mean x(F) =48.0+49.2+53.1+55.9+51.9+53.6+49.7+53.9+52.8+48.9/10

x(F) = 51.6

Now the standard deviation of financial planner is

SD(F) = √{∑(x-x(p))^2/n-1}

SD(F) = √(48.0-51.6)^2+(49.2-51.6)^2+..................+(48.9-51.6)^2/n-1

After solving;

SD(F) = 2.57

Test Statistic (t) = $\frac{x(P)-x(f)}{\sqrt{\left(\frac{(n(p)-1)s(p)^2+(n(f)-1)s(f)^2}{n(p)+n(f)-2}\left(\frac{1}{n(p)}+\frac{1}{n(f)}\right)\right)}}$

t = $\frac{55.2-51.61}{\sqrt{\left(\frac{(10-1)(3.34))^2+(10-1)(2.57)^2}{10+10-2}\left(\frac{1}{10}+\frac{1}{10}\right)\right)}}$

After solving

t = 2.6961

Hence, the value of the test statistic is 2.6961.

To learn more about test statistic link is here

brainly.com/question/14128303

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The right question is

public accountant 50.2 59.8 57.3 59.2 53.2 56.0 49.9 58.5 56.0 51.9

financial planner 48.0 49.2 53.1 55.9 51.9 53.6 49.7 53.9 52.8 48.9

Use a 0.05 level of significance and test the hypothesis that there is no difference between the starting annual salaries of public accountants and financial planner

Find the value of the test statistic.

4 0

1 year ago

Which of the following statements have the same result? Explain each step in solving each one.

STALIN [3.7K]

$f(x) = 5x + 1\\\\subtitute\ x = 1\ to\ the\ equation\\\\f(1) = 5\cdot1 + 1 = 5 + 1 = 6\\----------------------\\f(x)=\dfrac{2x+3}{5}\to y=\dfrac{2x+3}{5}\ \ \ \ |multiply\ both\ sides\ by\ 5\\\\5y=2x+3\ \ \ \ |subtract\ 3\ from\ both\ sides\\\\2x=5y-3\ \ \ \ \ |divide\ both\ sides\ by\ 2\\\\x=\dfrac{5y-3}{2}\\\\f^{-1}(x)=\dfrac{5\cdot(-1)-3}{2}=\dfrac{-5-3}{2}=\dfrac{-8}{2}=-4\\---------------------\\3y-7=y+5\ \ \ |add\ 7\ to\ both\ sides\\3y=y+12\ \ \ \ |subtract\ y\ from\ both\ sides\\2y=12\ \ \ \ |divide\ both\ sides\ by\ 2\\y=6$

Answer: f(1) when f(x) = 5x + 1 and 3y - 7 = y + 5

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3 years ago

A school has 360 students. 4/9 of them are girls. 3/5 of them are girls below the age of ten. How many girls are in the school a

quester [9]

To find total number of girls multiply total students by the fraction of girls:

360 x 4/9 = 360 x 4 = 1440/9 = 160 total girls.

If 3/5 are below the age of ten, that means 2/5 are above the age of ten.

Multiply the total number of girls by 2/5:

160 x 2/5 = 320/5 = 64

There are 64 girls above the age of ten.

4 0

3 years ago