1answer.
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
navik [9.2K]
3 years ago
10

PETS Becky wants to buy some fish for her aquarium. She has $20 to spend and the fish cost $2.50 each. Write and solve an inequa

lity to find how many fish Becky can afford.
Mathematics
1 answer:
Vilka [71]3 years ago
6 0
To start, you want to determine the inequality. You can do this by setting the number of buying n fish to be less than or equal to her current amount of money, being as it can't go over or she wouldn't be able to afford the fish. This inequality would be written so her amount of money (20) would be greater than or equal to the cost of n fish (2.5n), which would then look as such: 20≥2.5n. To solve this inequality, just solve for n by dividing both sides by 2.5, giving you 8≥n. This would mean that Becky can afford to buy up to 8 fish.
You might be interested in
How many fractions are equivalent to 4/5? explain
Pavel [41]
There are several fractions that are equivalent to 4/5. Some are 8/10, 80/100, 800/1000. They all equal the same decimal: 0.8
7 0
3 years ago
Hii giving thanks pls help
olchik [2.2K]

Answer:

x = 52

Step-by-step explanation:

<I = < K b/c of corresponding angles thm

x + 66 + 62 = 180

x + 128 = 180

x = 52

6 0
2 years ago
One end point of a line segment is ( -3, -6 ) The length of the line segment is 7 units find four points that could serve as the
Anika [276]

Answer:

(4, -6), (-10, -6), (-3, 1), (-3, -13)

Step-by-step explanation:

If we add/subtract t to/from any x/y coordinate, we will form a line segment with length 7. Hence, (4, -6), (-10, -6), (-3, 1), (-3, -13).

3 0
3 years ago
A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a s
RoseWind [281]

Answer:

There is a 95% confidence that the true mean height of all male student at the large college is between the interval (63.5, 74.4).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population mean is:

CI=\bar x\pm z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}

The (1 - α)% confidence interval for population parameter implies that there is a (1 - α) probability that the true value of the parameter is included in the interval.

Or, the (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the average height of male students at a large college is, (63.5 inches, 74.4 inches).

The 95% confidence interval for the average height of male students (63.5, 74.4) implies that, there is a 0.95 probability that the true mean height of all male student at the large college is between the interval (63.5, 74.4).

Or, there is a 95% confidence that the true mean height of all male student at the large college is between the interval (63.5, 74.4).

4 0
3 years ago
Please help 10 Points​
QveST [7]

Answer:

c

Step-by-step explanation:

8 0
3 years ago
Other questions:
  • Andrew is searching for a cup and discover that he has 20 plates for every 5 Pairs of cups if he has 40 plates how many pairs of
    9·1 answer
  • Is 84/45 a irrational repeating decimal or just a irrational one
    14·1 answer
  • 9x+12x-21=27<br> How do I solve ?
    9·1 answer
  • Which is a factor of 2x^2- 12x- 14
    7·1 answer
  • How to find the circumference of a circle
    7·1 answer
  • Question 7 (5 points)
    14·1 answer
  • 5/9 and 15/27 compare
    14·1 answer
  • Which of the following statements are true?<br><br><br> Please help..
    9·1 answer
  • Tell whether the angles are adjacent or vertical. Then find the value of x.
    14·1 answer
  • PLEASE HELP ITS MATH!!!!!
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!