​There are two types of fish in a lake. These are carp and pike.
In a netted area of the lake 120 carp and 16 pike were caught
In the whole lake it is estimated there are 34 000 fish.How​ many pike are there?​​

Answers

Answer 1

Answer:pike are 250 ; from 120+16=136 and 34000/136 is 250

Step-by-step explanation:


Related Questions

A tortoise is walking in the desert. It walks 7.5 meters in 3 minutes. What is its speed?

Answers

Answer:

Step-by-step explanation:

speed is calculated using formula v=d/t

m= 7.5m

t= 3 min

v=?

v= 7.5m/3min

v= 2.5m/min

HELP ASAP!!
The equation (blank) has no solution​

Answers

Answer:

Just to recap, an equation has no solution when it results in an incorrect "equation".

For  example:

Equation: x+3 = x+4

Subtract x: 3 = 4???

But clearly, 3 is not equal to 4, so this equation has NO SOLUTION.

Now onto our problem:

13y+2-2y = 10y+3-y

11y+2 = 9y+3

2y=1

y=1/2

9(3y+7)-2 = 3(-9y+9)

27y+61 = -27y+27

54y = -34

y = -34/54

32.1y+3.1+2.4y-8.2=34.5y-5.1

34.5-5.1=34.5y-5.1

5.1=5.1

infinite solutions

5(2.2y+3.4) = 5(y-2)+6y

11y+17 = 11y-10

17 = -10??

That's not true, so the option "5(2.2y+3.4) = 5(y-2)+6y" has no solution.

Let me know if this helps

The sum of the 3rd and 7th terms of an A.P. is 38, and the 9th term is 37. Find the A.P.

Answers

Answer:

The AP is 1, 11/2, 10, 29/2, 19, ....

Step-by-step explanation:

Let the first term be a and d be the common difference of the arithmetic progression.

ATQ, a+2d+a+6d=38, 2a+8d=38 and a+8d=37. Solving this, we will get a=1 and d=9/2. The AP is 1, 11/2, 10, 29/2, 19, ....

Suppose that 22 inches of wire costs 66 cents.
At the same rate, how much (in cents) will 17 inches of wire cost?
cents
Х
?

Answers

Answer:

51 cents for 17 inches of wire

Step-by-step explanation:

22 = 66

17 = x

22x = 66 * 17

22x = 1122

x = 51 cents

or

22 inches costs 66 cents

1 inch costs 3  cents (66 / 22 = 3  cents)

17 inches costs 51  cents (17 * 3 = 51 cents)

find the n^th root of z = -2i, n = 6​

Answers

Answer:

2^(1/6) (cos(-pi/12)+i sin(-pi/12))

2^(1/6) (cos(3pi/12)+i sin(3pi/12))

2^(1/6) (cos(7pi/12)+i sin(7pi/12))

2^(1/6) (cos(11pi/12)+i sin(11pi/12))

2^(1/6) (cos(5pi/4)+i sin(5pi/4))

2^(1/6) (cos(19pi/12)+i sin(19pi/12))

Step-by-step explanation:

Let's convert to polar form.

-2i=2(cos(A)+i sin(A) )

There is no real part so cos(A) has to be zero and since we want -2 and we already have 2 then we need sin(A)=-1 so let's choose A=-pi/2.

So z=2(cos(-pi/2)+i sin(-pi/2)).

There are actually infinitely many ways we can write this polar form which we will need.

z=2(cos(-pi/2+2pi k)+i sin(-pi/2+2pi k))

where k is an integer

Now let's find the 6 6th roots or z.

2^(1/6) (cos(-pi/12+2pi k/6)+i sin(-pi/12+2pi k/6))

Reducing

2^(1/6) (cos(-pi/12+pi k/3)+i sin(-pi/12+pi k/3))

Plug in k=0,1,2,3,4,5 to find the 6 6th roots.

k=0:

2^(1/6) (cos(-pi/12+pi (0)/3)+i sin(-pi/12+pi (0)/3))

=2^(1/6) (cos(-pi/12)+i sin(-pi/12))

k=1:

2^(1/6) (cos(-pi/12+pi/3)+i sin(-pi/12+pi/3))

2^(1/6) (cos(3pi/12)+i sin(3pi/12))

k=2:

2^(1/6) (cos(-pi/12+2pi/3)+i sin(-pi/12+2pi/3))

2^(1/6) (cos(7pi/12)+i sin(7pi/12))

k=3:

2^(1/6) (cos(-pi/12+3pi/3)+i sin(-pi/12+3pi/3))

2^(1/6) (cos(11pi/12)+i sin(11pi/12))

k=4:

2^(1/6) (cos(-pi/12+4pi/3)+i sin(-pi/12+4pi/3))

2^(1/6) (cos(15pi/12)+i sin(15pi/12))

2^(1/6) (cos(5pi/4)+i sin(5pi/4))

k=5:

2^(1/6) (cos(-pi/12+5pi/3)+i sin(-pi/12+5pi/3))

2^(1/6) (cos(19pi/12)+i sin(19pi/12))

I need to find the distance B in the special counter sink shown

Answers

Answer:

Step-by-step explanation:

87°32' = 86°92'

(86°92')/2 = 43°46'

B = 13/(16cos(43°46')) = 1.125

Answer:

Step-by-step explanation:

The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $440 and standard deviation $20. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1

Answers

Answer:

$465.6 should be budgeted.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean $440 and standard deviation $20.

This means that [tex]\mu = 440, \sigma = 20[/tex]

How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1?

The 100 - 10 = 90th percentile should be budgeted, which is X when Z has a p-value of 0.9, so X when Z = 1.28. Then

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.28 = \frac{X - 440}{20}[/tex]

[tex]X - 440 = 1.28*20[/tex]

[tex]X = 465.6[/tex]

$465.6 should be budgeted.

Which of the following inequalities matches the graph?
10
6
-10
Oxs-1
Ox>-1
Oys-1
Oy 2-1

Answers

Answer:

y > -1

Step-by-step explanation:

the line is going across the y axis and is everything above -1

Given the central angle, name the arc formed.
Major arc for ∠EQD

A. EQDˆ
B. GDFˆ
C. EGDˆ
D. EDˆ

Answers

9514 1404 393

Answer:

  C. EGD

Step-by-step explanation:

A major arc is typically named using the end points and a point on the arc. Here, the end points are E and D, and points on the major arc include C, G, and F. The major arc ED could be named any of

arc ECDarc EGD . . . . choice Carc EFD

Of course, the reverse of any of these names could also be used: DCE, DGE, DFE.

Find two power series solutions of the given differential equation about the ordinary point x = 0. Compare the series solutions with the solutions of the differential equation obtained using the method of Section 4.3. Try to explain any differences between the two forms of the solution. y'' − y' = 0 y1 = 1 − x2 2! + x4 4! − x6 6! + and y2 = x − x3 3! + x5 5! − x7 7! + y1 = x and y2 = 1 + x + x2 2! + x3 3! + y1 = 1 + x2 2! + x4 4! + x6 6! + and y2 = x + x3 3! + x5 5! + x7 7! + y1 = 1 + x and y2 = x2 2! + x3 3! + x4 4! + x5 5! + y1 = 1 and y2 = x + x2 2! + x3 3! + x4 4! +

Answers

You're looking for a solution in the form

[tex]y(x) = \displaystyle \sum_{n=0}^\infty a_nx^n[/tex]

Differentiating, we get

[tex]y'(x) = \displaystyle \sum_{n=0}^\infty na_nx^{n-1} = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]

[tex]y''(x) = \displaystyle \sum_{n=0}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=1}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n[/tex]

Substitute these for y' and y'' in the differential equation:

[tex]\displaystyle \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n - \sum_{n=0}^\infty (n+1)a_{n+1}x^n = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1)a_{n+2}-(n+1)a_{n+1}\bigg)x^n = 0[/tex]

Then the coefficients of y are given by the recurrence

[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\frac{a_{n+1}}{n+2}&\text{for }n\ge0\end{cases}[/tex]

or

[tex]a_n = \dfrac{a_{n-1}}n[/tex]

But we cannot assume that [tex]a_0[/tex] and [tex]a_1[/tex] depend on each other; we can only guarantee that the recurrence holds for n ≥ 1, so that

[tex]a_2=\dfrac{a_1}2 \\\\ a_3=\dfrac{a_2}3=\dfrac{a_1}{3\times2} \\\\ a_4=\dfrac{a_3}4=\dfrac{a_1}{4\times3\times2} \\\\ \vdots \\\\ a_n=\dfrac{a_1}{n!}[/tex]

So in the power series solution, we split off the constant term and we're left with

[tex]y(x) = a_0 + a_1 \displaystyle \sum_{n=1}^\infty \frac{x^n}{n!}[/tex]

so that the fundamental solutions are

[tex]y_1=1[/tex]

and

[tex]y_2=x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots[/tex]

Can someone help please

Answers

Step-by-step explanation:

a) The volume of the prism is

[tex]V = (n^2 - 1)×(n^2 - 1)×(5n)[/tex]

[tex]\:\:\:\:= (n^4 - 2n^2 + 1)(5n)[/tex]

[tex]\:\:\:\:=5n^5 - 10n^3 + 5n[/tex]

b) If the dimensions L of the prism are tripled, the new volume will be

[tex]V' = (3L)^3 = 27L^3 = 27V[/tex]

so it will increase by a factor of 27.

Team A scored 30 points less than four times the number of points that Team B scored. Team C scored 61 points more than half of the number of points that Team B scored. If Team A and Team C shared in the victory, having earned the same number of points, how many more points did each team have than Team B?

Answers

Answer:

team a and team c scored 74 points which is 48 points more than team b, scoring 26 points.

Step-by-step explanation:

Assume the random variable X is normally distributed with mean μ = 50 and standard deviation σ = 7. Compute the probability

P(35 < X < 58)= ________

Answers

Answer:

Probability-Between    .8574 = 85.74%

Step-by-step explanation:

Z1=-2.14 Z2=1.14

*x-1 35

*x-2  58

*µ 50

*σ 7

4. Explain how the graphs of the functions are similar and how they are different
2x+3y=1470
And
2x+3y=1593

Answers

Answer:

they are parallel lines so have the same slope.   Difference would be the where they intersect the x and y axis

Step-by-step explanation:

help pls! I need the answer quickly and pls explain. thank you!

Answers

Answer:

h = 6[tex]\sqrt{3}[/tex]

Step-by-step explanation:

The given is the special right triangle with angle measures : 90-60-30

and the side lengths for the given angles are represented by :

2a-a[tex]\sqrt{3}[/tex]-a

the side length that sees 60 degrees is represented by a[tex]\sqrt{3}[/tex] (h in this case)

the area of a triangle is calculated by multiplying height and base and that is divided by 2

a[tex]\sqrt{3}[/tex]*a/2 = 18[tex]\sqrt{3}[/tex] multiply both sides by 2

a^2[tex]\sqrt{3}[/tex] = 36[tex]\sqrt{3}[/tex] divide both sides by [tex]\sqrt{3}[/tex]

a^2 = 36 find the roots for both sides

a = 6

since h sees angle measure 60 and is represented by a[tex]\sqrt{3}[/tex]

h = 6[tex]\sqrt{3}[/tex]

If Bobby drinks 5 waters in 10 hours how many does he drink in 1 hour ?

Answers

Water drunk by Bobby in 10 hours = 5 units

So, water drink by Bobby in 1 hour

= 5/10 units

= 1/2 units

= 0.5 units

Answer:

1/2 water

Step-by-step explanation:

We can use a ratio to solve

5 waters        x waters

-----------   = ------------

10 hours             1 hours

Using cross products

5*1 = 10 *x

5 = 10x

Divide by 10

5/10 = x

1/2 waters =x

A student found the solution below for the given inequality.Which of the following explains whether the student is correct?The student is completely correct because the student correctly wrote and solved the compound inequality.The student is partially correct because only one part of the compound inequality is written correctly.The student is partially correct because the student should have written the statements using “or” instead of “and.”The student is completely incorrect because there is “ no solution “ to this inequality.

Answers

Answer:

The student is completely incorrect because there is no solution to this inequality.

Answer:

D on edge

Step-by-step explanation:

If it takes 5 years for an animal population to double, how many years will it take until the population
triples?

Answers

9514 1404 393

Answer:

  7.92 years

Step-by-step explanation:

We want to find t such that ...

  3 = 2^(t/5)

where 2^(t/5) is the annual multiplier when doubling time is 5 years.

Taking logs, we have ...

  log(3) = (t/5)log(2)

  t = 5·log(3)/log(2) ≈ 7.92 . . . years

It will take about 7.92 years for the population to triple.

Which of the following theorems verifies that AABC - ASTU?
A. AA
B. HL
C. HA
D. LL

Answers

Answer:

AA

Step-by-step explanation:

See In Triangle ABC and Triangle STU

[tex]\because\begin{cases}\sf \angle A=\angle S=90° \\ \sf \angle B=\angle T=31°\end{cases}[/tex]

Hence

[tex]\sf \Delta ABC~\Delta STU(Angle-Angle)[/tex]

By AA similarity  triangle ABC is similar to triangle SUT. Therefore, option A is the correct answer.

What are similar triangles?

Two triangles are similar if the angles are the same size or the corresponding sides are in the same ratio. Either of these conditions will prove two triangles are similar.

In the given triangle ABC, ∠C=180°-90°-31°

∠C=59°

In the given triangle SUT, ∠U=180°-90°-31°

∠U=59°

Here, ∠B=∠T (Given)

∠C=∠U (Obtained using angle sum property of a triangle)

So, by AA similarity ΔABC is similar to ΔSUT.

Therefore, option A is the correct answer.

To learn more about the similar triangles visit:

https://brainly.com/question/25882965.

#SPJ7

Each course at college X is worth either 2 or 3 credits. The members of the men's swim team are taking a total of 48 courses that are worth a total of 107 credits. How many 2-credit courses and how many 3-credit courses are being taken?

Answers

Answer:

Let the number of courses that are worth 3 credits each be x and those worth 4 credits be y. With the given information, you can write the following equations:  

x + y = 48

3x + 4y = 155

 

You can solve the above equations by method of elimination/substitution

x + y = 48 ⇒ x = 48 - y (Now, substitution this equation into 3x + 4y = 155)

3(48 - y) + 4y = 155

144 -3y + 4y = 155

y + 144 = 155

y = 11

 

Now plug this solution back into x = 48 - y  

x = 48 - 11 = 37  

 

Check work (by plugging the solutions back into the 3x + 4y and see if it's equal to 155):

3(37) + 4(11) = 155

Answer: There are 37 of the 3-credit course and 11 of the 4-credit course

Plz help a beggar I don’t get it

Answers

Answer: 3

happy learning

Answer:

B.

Step-by-step explanation:

From the point (-1,0) the next point on the graph is up 3, right 1, making the slope a positive 3.

Follow the instructions on the image

Answers

Answer:

k=3

Step-by-step explanation:

Assuming the centre of dilation is 0,0, we can use the formula (kx,ky) to determine it.

Here,

The co-ordinates of pre-image=(0,1),(-1,-1) & (1,-1)

The co-ordinates of image=(0,3),(-3,-3) & (3,-3)

Now,

(kx,ky)=(0,3)

(k*0,k*1)=(0,3)

Equating,

k=3

You can use the other coordinates to further solidify your answer.

Find the arclength of the curve r(t) = ⟨ 10sqrt(2)t , e^(10t) , e^(−10t)⟩, 0≤t≤1

Answers

Answer:

[tex]\displaystyle AL = 2sinh(10)[/tex]

General Formulas and Concepts:

Pre-Calculus

Hyperbolic Functions

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Derivative Rule [Chain Rule]:                                                                                 [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Exponential Differentiation

Integration

IntegralsIntegration Constant CDefinite Integrals

Parametric Integration

Vector Value Functions

Vector Integration

Arc Length Formula [Vector]:                                                                               [tex]\displaystyle AL = \int\limits^b_a {\sqrt{[i'(t)]^2 + [j'(t)]^2 + [k'(t)]^2}} \, dt[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \vec{r}(t) = <10\sqrt{2}t , e^{10t} , e^{-10t} >[/tex]

Interval [0, 1]

Step 2: Find Arclength

Rewrite vector value function:                                                                     [tex]\displaystyle r(t) = 10\sqrt{2}t \textbf i + e^{10t} \textbf j + e^{-10t} \textbf k[/tex]Substitute in variables [Arc Length Formula - Vector]:                               [tex]\displaystyle AL = \int\limits^1_0 {\sqrt{\bigg[\frac{d}{dt}[10\sqrt{2}t \textbf i]\bigg]^2 + \bigg[\frac{d}{dt}[e^{10t} \textbf j]\bigg]^2 + \bigg[\frac{d}{dt}[e^{-10t} \textbf k ]\bigg]^2}} \, dt[/tex][Integrand] Differentiate [Respective Differentiation Rules]:                     [tex]\displaystyle AL = \int\limits^1_0 {\sqrt{[10\sqrt{2} \textbf i]^2 + [10e^{10t} \textbf j]^2 + [-10e^{-10t} \textbf k]^2}} \, dt[/tex][Integrand] Simplify:                                                                                       [tex]\displaystyle AL = \int\limits^1_0 {\sqrt{200 \textbf i + 100e^{20x} \textbf j + 100e^{-20x} \textbf k}} \, dt[/tex][Integral] Evaluate:                                                                                         [tex]\displaystyle AL = 2sinh(10)[/tex]

Topic: AP Calculus BC (Calculus I + II)

Unit: Vector Value Functions

Book: College Calculus 10e

100.331 divide 99.355​

Answers

Answer:

1.009823361

Step-by-step explanation:

Just divide like this:

[tex] \frac{100.331}{99.355} = 1.009853361[/tex]

Fourteen boys and 21 girls will be equally divided into groups. Find the greatest number of groups that can be created if no one is left out.

Answers

Answer:

7 groups can be made each with five people :)

f=((-1,1),(1,-2),(3,-4)) g=((5,0),(-3,4),(1,1),(-4,1)) find (fg)(1)

Answers

Answer:

f(g(1)) = - 2

Step-by-step explanation:

Find g(1) then use the value obtained to find f(x)

g(1) = 1 ← value of y when x = 1 (1, 1 ) , then

f(1) = - 2 ← value of y when x = 1 (1, - 2 )

A telescope contains both a parabolic mirror and a hyperbolic mirror. They share focus ​, which is 46feet above the vertex of the parabola. The​ hyperbola's second focus is 6 ft above the​ parabola's vertex. The vertex of the hyperbolic mirror is 3 ft below . Find the equation of the hyperbola if the center is at the origin of a coordinate system and the foci are on the​ y-axis. Complete the equation.

Answers

the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.

The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.

In terms of hyperbola, F1F2=2c, c=20.

The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.

Use formula c^2=a^2+b^2c

2

=a

2

+b

2

to find b:

\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}

(20)

2

=(18)

2

+b

2

,

b

2

=400−324=76

.

The branches of hyperbola go in y-direction, so the equation of hyperbola is

\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1

b

2

y

2

a

2

x

2

=1 .

Substitute a and b:

\dfrac{y^2}{76}- \dfrac{x^2}{324}=1

76

y

2

324

x

2

=1 .

Two complementary angles have measures of s and t. if t is less than twice s, which system of linear equations can be used to determine the measure of each angle? Please explain answer. I know s+t=90. But how do you get to t=2s-90

Answers

The required expressions are both equations 1 and 2 as shown:

[tex]s + t = 90 ......... 1[/tex]

[tex]t<2s[/tex] .... 2

Complementary angles are angles that sum up to 90 degrees. For instance, and A and B are complementary if A + B = 90.

According to the question, if two complementary angles have measures of s and t then:

[tex]s + t = 90 ......... 1[/tex]

Twice of 's' is expressed as [tex]2s[/tex]

If t is less than twice s, this can be expressed as [tex]t<2s[/tex] .... 2

The required expressions are both equations 1 and 2 as shown:

[tex]s + t = 90 ......... 1[/tex]

[tex]t<2s[/tex] .... 2

Learn more on word problems leading to simultaneous equations here: https://brainly.com/question/14294864

find the measure of d

Answers

You’re looking at an 90 degree angle. So you would just subtract 14 degrees from 90 degrees. Your answer is 76 degrees

After leaving an airport, a plane flies for 2 hours on a course of 60 degrees at a speed of 200 kilometers per hour. The plane then flies for 3 hours on a course of 210 degrees at a speed of 100 kilometers per hour What is the distance of the airport from the plane in kilometers? Round to the nearest tenth​

Answers

Answer: 205.3

I suppose all measures of angles are done from the same axis (for example x-axis)

Step-by-step explanation:

You just have to use the theorem of Al'Kashi:

[tex]d^2=400^2+300^2-2*300*400*cos(30^o)\\\\d\approx{205.3(km)}[/tex]

Other Questions
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