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3 years ago

Write a world problem that might come up in your everyday life. Solve the problem and pass right for your partner to solve

Mathematics

1 answer:

sammy [17]3 years ago

4 0

Answer:

This is a real life word problem about me lol.

Seth drinks half a gallon of milk a day, how many cups is this?

For one, we have to know how many quartz are in half a gallon, there are 2.

There are also 4 cups in a quartz, half a gallon is 2 quartz.

4*2=8

Seth drinks 8 cups of milk per day.

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Create your own scenario for a system of equations and solve it. Please do not just write two equations, but apply what you have

BARSIC [14]

Answer:

Dr. Potter provides vaccinations against polio and measles.

Each polio vaccination consists of 444 doses, and each measles vaccination consists of 222 doses. Last year, Dr. Potter gave a total of 606060 vaccinations that consisted of a total of 184184184 doses.

How many polio vaccinations and how many measles vaccinations did Dr. Potter give last year?

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The average human body contains about 40 L of water. What is the total amount of water contained in the 6.4 billion people in th

vesna_86 [32]

Answer:

2.56 * 10^11 L

Step-by-step explanation:

1 body has 40 L

Thus, the amount in 6.4 billion would be;

40 * 6.4 * 10^9

= 256 * 10^9

= 2.56 * 10^2 * 10^9

= 2.56 * 10^11 L

7 0

3 years ago

Last year, 85% of students participated in the school fundraiser. Using this data, what would be the BEST prediction of how many

SpyIntel [72]

Answer:

C. 272

Step-by-step explanation:

Since we know 85% of the students participated last year we can assume the same amount will participate this year so to find the answer you simply do the total number of students (320) multiplied by 85% or .85 to get the best guess of how many students will participate this year.

8 0

3 years ago

Read 2 more answers

Help me please! BRAINLIEST

masya89 [10]

Answer:

1. 53 by 58 , 222 ft

2. 321000

3.1080

4. 3.5

Step-by-step explanation:

1. 23+15+15 = 53

28+15+15= 58

53+53+58+58= 222

6 0

3 years ago

A group of 121 Internet users 50-65 years of age were randomly chosen and asked to report the weekly number of hours they spend

Luda [366]

Answer: We support the null hypothesis.

Step-by-step explanation:

Given : The purpose of the study was to determine whether the mean weekly number of hours that Internet users in that age group spend online differs from the mean for Internet users in general which is 12.50 hours.

The set of hypothesis for the given claim :-

$H_0:\mu=12.50\\\\ H_a:\mu\neq12.50$

A 95% confidence interval for the mean weekly number of hours that Internet users in that age group spend online was calculated to be (11.439 , 12.577).

Since 12.50 contained in the 95% confidence interval (11.439 , 12.577).

Therefore, we support the null hypothesis.

So the final conclusion will be, we have evidence to support the claim that Internet users in that age group spend online differs from the mean for Internet users in general which is 12.50 hours.

7 0

4 years ago