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
zzz [600]
3 years ago
7

Marc showed 7 with this addition sentence 7=3+3+1. Use marc’s way to show these odd numbers with addition sentences . 5=

Mathematics
1 answer:
Doss [256]3 years ago
3 0
The answer is: 5=3+1+1
You might be interested in
30% of the toys colleted at the holiday toy drive were stuffed animals.There were a total of 80 toys collected.How many toys wer
xxTIMURxx [149]

Answer:

24

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What percent of 18 is 36
enot [183]
So you just want to say 36/18 or 2.

So it's 2, or 200%.
7 0
3 years ago
Read 2 more answers
100 POINTS & BRAINLIEST!
Kipish [7]
The answers are D&E hope this helped !
8 0
3 years ago
The five members of the Traynor family each buy train tickets. During the train ride, each family member buys a boxed lunch for
larisa [96]

Let's assume

the price of each train ticket is x

we are given

total number of family members =5

so, total cost of ticket = total number of family members*the price of each ticket

so, total cost of ticket =5*x

so, total cost of ticket =5x

now, we have

each family member buys a boxed lunch for $6.50

so, total cost of lunch box = total number of family members*price of each lunch box

so, total cost of lunch box = 5*6.50

total cost of trip = total cost of ticket+total cost of lunch box

we are given

total cost of trip is $248.50

now, we can plug values

and we get

248.50=5x+5*6.50

now, we can solve for x

248.50=5x+32.5

248.50-32.5=5x+32.5-32.5

216=5x

\frac{5x}{5} =\frac{216}{5}

x=43.2

so,

the price of each train ticket is $43.2..........Answer

8 0
3 years ago
Read 2 more answers
The following probabilities are based on data collected from U.S. adults during the National Health Interview Survey 2005-2007.
Nutka1998 [239]

Answer:

Required Probability  =0.421  

Step-by-step explanation:

Let's first arrange the data given in a more presentable way: So, we have the following probabilities for different categories.

Underweight (UW) = 0.019

Healthy Weight (HW) = 0.377

Overweight but Not Obese (NO) = 0.35

Obese (O) = 0.254

<em>Now, let's calculate the probability that a randomly selected American adult who weighs more than the healthy weight range is obese:</em>

Required Probability = Probability(obese)/Probability(Overweight + Obese)

                                    = P (O) /P(NO + O)                                  

                                    =0.254/(0.35+0.254)

Required Probability  =0.421

where, O for Obese and NO for Not Obese or Overweight but Not Obese.

So, the correct answer = 0.421

8 0
3 years ago
Other questions:
  • Sara loves to go hiking. this week she hiked 3 miles on monday, 4 miles on tuesday, and 3 miles on thursday. last week sarah hik
    5·2 answers
  • I need help on this math problem help !?
    8·2 answers
  • Find the mean, median, mode, and range of these numbers 75,95,90,95,60,95,75,95,90<br> plz
    14·1 answer
  • Dustin makes $2,330 each month and pays $840 for rent. To the nearest tenth of a percent, what percent of Dustin's earning are s
    7·1 answer
  • Jeffrey has $250 in the bank.Each week he takes out $25.After a few weeks jeffrey has $87.50 left in the bank. What is the numbe
    9·1 answer
  • Cory paid $9 for lunch and paid $11 to go to a movie. How much money did he spend?
    8·1 answer
  • What are the values of x, y, and z? Use theorems to justify each answer.<br>​
    12·2 answers
  • What is the equation of the line that passes through the point (5, - 7) and has a slope of o ?
    14·1 answer
  • PLEASE SEND HELP WE DO NOT KNOW HOW TO DO THIS
    15·1 answer
  • PLEASE HELP IM REALLY STUCK
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!