answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
valentina_108 [34]
2 years ago
12

A group of entomologists has determined that the population of ladybugs at a local park can be modeled by the equation y = − 1.4

37 x + 197.686 , where x represents the number of years since 2010 and y represents the number of ladybugs, in thousands.
a) Predict the ladybug population at the park in 2024.
b) Predict the ladybug population at the park in 2060.
Mathematics
1 answer:
Oksanka [162]2 years ago
7 0
<h3>Answer:</h3>

A) 177.568 thousand.

B) 125.836 thousand.

<h3>Step-by-step explanation:</h3>

In this question, it is asking you to use the equation to find the population of ladybugs in a certain year.

Equation we're going to use:

y = -1.437 x + 197.686

We know that the "x" variable represents the number of years since 2010, so that means our starting year is 2010.

Lets solve the question.

Question A:

We need to find the ladybug population is 2024.

2024 is 14 years after 2010, so our "x" variable will be replaced with 14.

Your equation should look like this:

y = -1.437 (14) + 197.686

Now, we solve.

y = -1.437 (14) + 197.686\\\\\text{Multiply -1.437 and 14}\\\\y=-20.118+197.686\\\\\text{Add}\\\\y=177.568

You should get 177.568

This means that the population of ladybugs in 2024 is 177.568 thousand.

Question B:

We need to find the ladybug population is 2060.

2060 is 50 years after 2010, so the "x" variable would be replaced with 50.

Your equation should look like this:

y = -1.437 (50) + 197.686

Now, we solve.

y = -1.437 (50) + 197.686\\\\\text{Multiply -1.437 and 50}\\\\y=-71.85+197.686\\\\\text{Add}\\\\y=125.836

This means that the population of ladybugs in 2060 would be 125.836 thousand.

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>
You might be interested in
7, 16, 25, 34, 43, ..... nth term is
vladimir2022 [97]

Answer:

79 is the ninth term.

Step-by-step explanation:

You add nine to each one.

7, 16, 25, 34, 43, 52, 61, 70, 79

79 is the ninth term.

6 0
2 years ago
Read 2 more answers
Let p, q, and r be the propositions p: You get an A on the final exam. q: You do every exercise in this book. r: You get an A in
babunello [35]

Answer:

a) r ^ ¬q

b) p ^ꓥqꓥ^ r

c) r → p

d) p ^ ¬qꓥ^ r

e) (p ^ q) → r

f) r ↔ (q v p)

Step-by-step explanation:

 

 

ꓥ^ = AND Conjunction

vꓦ = OR disjunction

¬ = NOT Negation

↔ = Double Implication

→ = implication

6 0
1 year ago
What is the sum of square root of -2 and square root of -18
Tasya [4]
Answer is B
4 square root of 2i

7 0
1 year ago
Rearrange x=3g+2 to make g the subject
Eduardwww [97]

Answer:

g = (x - 2)/3

Step-by-step explanation:

g = (x - 2)/3

8 0
1 year ago
Read 2 more answers
A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave
erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, L = 120 cm

Its wavelength \lambda_1 = 480 cm

                         \lambda_2 = 160 cm and \lambda_3 = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

  • The fundamental wavelength of a pipe which is open at both ends is 2L.
  • The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ 4L=4(120)=480\ cm

It seems that the pipe is open at one end and closed at one end.

Now lets check with the subsequent wavelengths.

For one side open and one side closed pipe:

An odd-integer number of quarter wavelength have to fit into the tube of length L.

⇒  \lambda_2=\frac{4L}{3}                                   ⇒  \lambda_3=\frac{4L}{5}

⇒ \lambda_2=\frac{4(120)}{3}                              ⇒  \lambda_3=\frac{4(120)}{5}

⇒ \lambda_2=\frac{480}{3}                                  ⇒  \lambda_3=\frac{480}{5}

⇒ \lambda_2=160\ cm                           ⇒   \lambda_3=96\ cm  

So the pipe is open at one end and closed at one end .

6 0
1 year ago
Other questions:
  • Which values of c will cause the quadratic equation –x2 + 3x + c = 0 to have no real number solutions? Check all that apply.
    12·2 answers
  • When Jennifer spent 3/7 of her money, she has $72 left. How much money did Jennifer start with?
    15·1 answer
  • Mr. Carandang sold a total of 1,790 prints of one of his drawings. Out of all 1,273 unframed prints that he sold, 152 were small
    12·2 answers
  • a survey asked 50 students if they play an instrument and if they are in band. 25 students play an instrument 20 are in a band a
    14·2 answers
  • Which square root is a whole number? A) 195 B) 196 C) 197 D) 198
    7·2 answers
  • Identify the domain of the function shown in the graph
    9·2 answers
  • In Lixue's garden, the green pepper plants grew 5 inches in 3/4 month. At this rate, how many feet can they grow in one month? (
    15·1 answer
  • Sofia cuts a piece of felt in the shape of a kite for an art project. The top two sides measure 20 cm each and the bottom two si
    13·2 answers
  • Floors, Inc. offers terms of 2/10, n/30 to credit customers. Tile Magic Corp. purchased 100 tile cutters with a list price of $2
    9·1 answer
  • A department store purchases screen-printed t-shirts at a cost of $5 per shirt. They mark up the price 150% (making the selling
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!