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horsena [70]
3 years ago
7

Phone calls cost $7 for 3 minutes how many dollars per minute

Mathematics
1 answer:
Ilya [14]3 years ago
3 0
7/3 divide up 7 dollars into 3 minutes by typing this into a calculator
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What is the greatest common factor between 72 and 54?
Elanso [62]
The greatest common factor between 72 and 54 is 18.
3 0
3 years ago
A box of oat cereal costs $3.90 for 15 ounces. A box of rice of cereal costs $3.30 for 11 ounces. Which box of cereal costs less
Firlakuza [10]

Answer:

The cost of per ounce of oat cereal is $0.04 less than per ounce cost of rice.

Step-by-step explanation:

Given:

Cost of 15 ounces of oat cereal box = $3.90

Cost of 11 ounces of rice box = $3.30

To compare the unit prices of the the boxes.

Solution:

Using unitary method to find the unit prices:

If 15 ounce of oat box cost =$3.90

∴ 1 ounce of oat cereal would cost = \$\frac{3.90}{15}=\$0.26

∴ Per ounce cost of oat cereal = $0.26

If 11 ounce of rice box cost =$3.30

∴ 1 ounce of rice would cost = \$\frac{3.30}{11}=\$0.3

∴ Per ounce cost of rice = $0.3

Difference in unit costs of oat and rice = \$0.3-\$0.26=\$0.04

On comparing the the unit costs of oats and rice, we find out that the cost of per ounce of oat cereal is $0.04 less than per ounce cost of rice.

3 0
3 years ago
How do you solve this absolute value equatuon? 4|x - 1| - 7 = -3
MatroZZZ [7]
4|x - 1| - 7 = -3    |add 7 to both sides

4|x - 1| = 4    |divide both sides by 4

|x - 1| = 1 ⇔ x - 1 = 1 or x - 1 = -1    |add 1 to both sides

x = 2 or x = 0

Answer: B.
3 0
3 years ago
A taxi driver charges a fixed rate of r to pick up a passenger. In addition, the taxi driver charges a rate of m for each mile d
aivan3 [116]
T= MN+R
I believe that this is the right answer
8 0
3 years ago
John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
2 years ago
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