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Georgia [21]
3 years ago
15

In a certain town 60% of the households own mutual funds, 40% own individual stocks, and 20% own both mutual funds and individua

l stocks. The proportion of households that own mutual funds but not individual stocks is:A) 20%.
B) 30%.
C) 40%.
D) 50%

Mathematics
1 answer:
Genrish500 [490]3 years ago
6 0

Answer:

C. 40%

Step-by-step explanation:

Using set notations,

Check the attachment for the diagram.

Let the total fund shared by the town be 100% which will be our universal set.

Let X be proportion of households that own mutual funds but not individual

From the venn diagram,  

The total number of people that owned mutual fund M = (proportion of  households that owned both mutual fund and individual stock) +  (proportion of households that own mutual funds but not individual stocks)

If X is the proportion of households that own mutual funds but not individual stocks

The total number of people that owned mutual fund = (proportion of  households that owned both mutual fund and individual stock) +  X

X = (the total number of people that owned mutual fund)- (proportion of  households that owned both mutual fund and individual stock)

X = 60% - 20%

X = 40%

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Answer:

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Step-by-step explanation:

Given

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4 0
3 years ago
How many sets of three consecutive integers are there in which the sum of the three integers equals their product?
skad [1K]

Answer:

3

Step-by-step explanation:

since the 3 integers are consecutive, we are dealing with x, x+1, x+2.

and their sum is the same as their product :

x + (x + 1) + (x + 2) = x(x + 1)(x + 2)

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this is a polynomial of third degree.

and as such it has 3 solutions.

of course, it could be that some of them are the same or are even in the realm of complex numbers (i = sqrt(-1)), but usually these 3 solutions are different real numbers.

I tried x=1 just to see, and, hey, it is a solution for this equation.

x = 1 means that the other 2 consecutive integers are 2 and 3.

and indeed, 1+2+3 = 1×2×3 = 6.

now it is easier to find the other 2 solutions, as a zero solution can be expressed as a factor of the whole expression.

for x = 1 the factor term is (x - 1), as this term is then turning 0, when x = 1.

I can divide the main expression by this factor and then analyze the quotient about the other 2 solutions.

x³ + 3x² - x - 3 : x - 1 = x² + 4x + 3

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----------------

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0 0

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the general solution to a quadratic equation is

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in our case

a = 1

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x = (-4 ± sqrt(4² - 4×1×3))/(2×1) =

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so, we have the additional solutions :

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and there we have it fully proven :

there are 3 different sets of 3 consecutive integers with the same sum as product.

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