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MArishka [77]
4 years ago
13

While traveling to Europe, Phelan exchanged 250 US dollars for euros. He spent 150 euros on his trip. After returning to the Uni

ted States he converts his money back to US dollars. How much of the original 250 US dollars does Phelan now have?
Mathematics
2 answers:
Anettt [7]4 years ago
5 0
Since the exchange rate of euro to USD right now is 1:1.11, therefore he spent 150x1.11=166.5 USD. Therefore, he has 250-166.5=83.5 USD left(assuming the exchange rate is the same, 1.17).
Lesechka [4]4 years ago
4 0

Answer:

Phelan have 80.5 US dollars.

Step-by-step explanation:

The current exchange rate is

1 \text{ Euro }=1.13 \text{ US dollars}

Let as assume that this exchange rate remains constant.

In initial stage Phelan has 250 US.

He spent 150 euros, it means he spent

150\times 1.13=169.5\text{ US dollars}

The remaining amount in US dollars is

250-169.5=80.5

Therefore, Phelan have 80.5 US dollars.

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Shelia went snowboarding. She spent $118 for the pass and snowboard rental. The snowboard rental cost $25. What percent of the t
Gnesinka [82]

Answer:

82.52%

Step-by-step explanation:

118+25=143

\frac{118}{143} =\frac{x}{100}

solve for x

(118*100)/143= 82.5175

4 0
3 years ago
Shenhed finished his math assament in 3/8
Elenna [48]
Math assignment = 3/8 hours
History assignment = 1/4 hours

We can not compare this as both fractions do not have alike denominators.

We can fix this problem easily by turning 3/8 and 1/4 into alike denominator fractions.

We can turn 1/4 into 2/8 as it is equivalent fractions.

1 x 2 = 2
4 x 2 = 8

(Whatever you do to the denominator, you have to do the same to the numerator)

Once we get this, we can see that Shen spent 2/8 hours on his homework while he spent 3/8 hours on his math assignment.

3/8 - 2/8 = 1/8

1/8 = 0.125

Shen spent 1/8 of an hour (or 0.125 of an hour) more on his math assignment than his history assignment.
4 0
2 years ago
A breed of cattle needs at least 10 proteins and 8 fats units per day. Feed type I provides 6 protein and 2 fat units at $4 per
Vikki [24]
Let the number of bags of feed type I to be used be x and the number of bags of feed type II to be used be y, then:

We are to minimize:

C = 4x + 3y

subject to the following constraints:

6x+2y\geq10 \\  \\ 2x+3y\geq8 \\  \\ x\geq0,\ y\geq0

From the graph of the 4 constraints above, the corner points are (0, 5), (1, 2), (4, 0).

Testing the objective function for the minimum corner point we have:

For (0, 5):
C = 4(0) + 3(5) = $15

For (1, 2):
C = 4(1) + 3(2) = 4 + 6 = $10

For (4, 0):
C = 4(4) + 3(0) = $16.

Therefore, the combination that yields the minimum cost is 1 bag of type I feed and 2 bags of type II feed.
7 0
3 years ago
B) Re-write the statement 16 + 5 + 2 - 1 by including one pair of brackets
Furkat [3]
B) 16 + (5 + 2) - 1

I don’t know what C is
3 0
3 years ago
2x^2 / x^2 - y^2 - x^2 / xy - y ^ 2​
Juliette [100K]

Answer:

2 - 2y^2 - x/y

Step-by-step explanation:

2x^2/x^2 = 2

and

x^2/xy = x/y

therefore

2x^2/x^2 - y^2 - x^2/xy - y^2

= 2 - y^2 - x/y - y^2

= 2 - y^2 - y^2 - x/y

= 2 - 2y^2 - x/y

6 0
3 years ago
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