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Leto [7]
3 years ago
11

At a supermarket in France, the price of apples is 2.50 euros per kilogram. Suppose the exchange rate is 1 euro =$1.34. Whoa is

the price of th apples in dollars per pound ?
Mathematics
1 answer:
kolezko [41]3 years ago
8 0
Here is the solution based on the given problem above:
2.5 euros = price of apples per kilo
1 euro = $1.34
$3.35 = price of apples per kilo
1 kilo = 2.2 pounds
Given that it is $3.35 per 2.2 pounds, so per pound, the price would be $0.66.
The correct answer would be $0.66/lb.

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Solve the inequality c + 6 < -20
bulgar [2K]

Answer:

The answer is c<−26

Step-by-step explanation:

7 0
3 years ago
You have $10,000 to invest in a stock portfolio. Your choices are Stock X with an expected return of 13 percent and Stock Y with
Alex Ar [27]

The money invested in stock X is $8,800.

Given,

Total investment in stock portfolio=$10000

Expected return from stock X=13%=0.13

Expected return from stock Y=8%=0.08

Expected return from portfolio=12.4%=0.124

The portfolio return is calculated by taking the weighted average of individual stock returns.

Let x represent the portfolio weightage of stock X.

1-x is the portfolio weightage of stock Y.

money invested in stock X be

0.124=0.13x+0.08(1-x)

0.124=0.13x+0.08-0.08x

0.124-0.08=0.05x

0.044=0.05x

x=0.88

money invested in stock X=0.88*10000=$8800

Thus, the money invested in stock X is $8,800.

To learn more about percentages refer here

brainly.com/question/26718419

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4 0
1 year ago
4 x 2.36 step by step
tensa zangetsu [6.8K]
9.44
4*2.36
All you would do is multiply 4 by 2.36 to get the answer.
4 0
3 years ago
Find the indicated function for f(x) = x^2 + 3x -1 and g(x) = 3x . Question <br>1. (f +g) (x) =
vesna_86 [32]

Answer:

Step-by-step explanation:

x^2 + 3x - 1 + 3x

x^2 + 6x - 1 is the solution

8 0
4 years ago
1/(x-1)x+1/x(x+1)+...+1/(x+9)(x+10)=11/12
nata0808 [166]

Notice that

\dfrac1{(x-1)x} = \dfrac1{x-1} - \dfrac1x \\\\ \dfrac1{x(x+1)} = \dfrac1x - \dfrac1{x+1} \\\\ \vdots \\\\ \dfrac1{(x+9)(x+10)} = \dfrac1{x+9} - \dfrac1{x+10}

That is, the <em>n</em>-th term (where <em>n</em> = -1, 0, 1, …, 9) in the sum on the left side has a partial fraction decomposition of

\dfrac{1}{(x+n)(x+n+1)} = \dfrac1{x+n} - \dfrac1{x+n+1}

and in the sum, some adjacent terms will cancel and leave you with

\dfrac1{(x-1)x} + \dfrac1{x(x+1)} + \cdots + \dfrac1{(x+9)(x+10)} = \dfrac1{x-1} - \dfrac1{x+10} = \dfrac{11}{12}

Now solve for <em>x</em>, bearing in mind that we cannot have <em>x</em> = 0, -1, -2, …, -10 :

\dfrac1{x-1} - \dfrac1{x+10} = \dfrac{11}{12}

Combine the fractions on the left side:

\dfrac{(x+10)-(x-1)}{(x-1)(x+10)} = \dfrac{11}{(x-1)(x+10)} = \dfrac{11}{12}

Then we must have

(x-1)(x+10) = x^2 + 9x - 10 = 12 \implies x^2+9x-22 = (x+11)(x-2) = 0

so that either <em>x</em> = 2 or <em>x</em> = -11.

4 0
3 years ago
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