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romanna [79]
2 years ago
6

landon owns a hybrid SUV that can travel 400 miles on a 15-gallon tank of gas.Determine how many miles he can travel on a 6 gall

ons.
Mathematics
2 answers:
Vedmedyk [2.9K]2 years ago
7 0
From another brainly answer "There are few ways to count it. One of them is:

400 miles on 15-gallon tank
400 / 15 = 26,66666666666667 miles per 1 gallon (looks weirdly but we'll get a nice result in a second)
So 26,66666666666667 * 6 gallons = 160 miles

<span>Answer: Landon's SUV can travel </span>160 miles<span> on 6 gallons."</span><span />
elixir [45]2 years ago
7 0
Let x be miles 400miles/15gl=x/6gl Cross multiplication (X)15gl=400miles(6gl) 15x=2400miles X=2400miles/15 X=160 miles
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Melanies fruit punch recipe calls for 6 cups of apple juice for every 3 cups os carrot juice how many cups os carrot juice are u
olchik [2.2K]
Well 3 is half of 6, so half of one is 1/5
5 0
2 years ago
Shelley receives supplies for her sno-cones business weekly. She usually works concession at four events a week and averages sal
My name is Ann [436]

Answer:

<u>Shelley did not budget enough. She will need additional US$ 16 for fulfilling the average sales of 760 sno-cones at the 4 events. With the budget she has, she just can supply 714 sno-cones (250 / 0.35).</u>

Step-by-step explanation:

1. Let's check all the information given for solving this question:

Cost of individual supply = US$ 0.35

Budget per week = US$ 250

Average sales of sno-cones per event = 190

Number of events per week = 4

2. Let's check if Shelley budgeted enough for the week events:

Total of sno-cones Shelley will sell = Average sales * Number of Events

Total of sno-cones Shelley will sell = 190 * 4 = 760

Cost of supplies for the week = Total of sno-cones * Cost of individual supply

Cost of supplies for the week = 760 * 0.35 = US$ 266

Budget per week = US$ 250

Budget per week  - Cost of supplies for the week = 250 - 266 = - 16

<u>Shelley did not budget enough. She will need additional US$ 16 for fulfilling the average sales of 760 sno-cones at the 4 events. With the budget she has, she just can supply 714 sno-cones (250 / 0.35).</u>

4 0
2 years ago
If one lens increases the size of an image by $80\%$ and another increases the size by $50\%$ by what percent will the image be
son4ous [18]

Answer:

Step-by-step explanation

Look at it this way:

If you increase something by 80 percent, your multiplying it by 1.8. This goes the same way for 50 percent. If you multiply 1.8 by 1.5, you will get 2.7. Then convert 2.7 to increasing something by 170 percent. So the answer is 170 percent.

8 0
2 years ago
Read 2 more answers
Derrick needs to figure out how he’s doing on his test scores so far this year. You can help by calculating the mean and the med
romanna [79]

Answer:

74.4 ,81.5 , median

Step-by-step explanation:

ed20 answers

4 0
2 years ago
Read 2 more answers
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
2 years ago
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