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2 years ago

6

Information was collected on those who attended the opening of a new movie. The analysis found that 56 percent of the moviegoers

were female, 26 percent were under age 25, and 17 percent were females under the age of 25. Find the probability that a moviegoer is either female or under age 25.

1 answer:

kherson [118]2 years ago

6 0

Answer: 0.65

Step-by-step explanation:

Let A denote that moviegoers were female and B denotes that moviegoers were under age 25.

As per given we have,

P(A)=0.56 P(B)=0.26 P(A∩B)=0.17

Using formula ,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\ P(A\cup B)=0.56+0.26-0.17=0.65$

Hence, the probability that a moviegoer is either female or under age 25 = 0.65

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2 years ago

The table below shows the amount paid for different numbers of items. Determine if this relationship forms a direct variation. V

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<h2>Answer/Step-by-step explanation:</h2>

Direct variation occurs when a variable varies directly with another variable. That is, as the x-variable increases, the y-variable also increases.

The ratio of between y-variable and x-variable would be constant.

Direct variation can be represented by the equation, $y = xk$ , where k is a constant. Thus,

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From the table given, it seems, as x increases, y also increases. Let's find out if there is a constant of proportionality (k).

Thus, ratio of y to x, $\frac{0.50}{1} = 0.5$

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If the given table of values has a direct variation relationship, then, plugging in the values of any (x, y), into $\frac{y}{x} = k$ , should give us the same constant if proportionality.

Let's check:

When x = 2, and y = 1:

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$\frac{1.5}{3} = 0.5$ ,

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$\frac{2.5}{5} = 0.5$ ,

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Answer:

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b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

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Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

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A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

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which can be reduced to (π + 4)y = 240 by dividing through with 2

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∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

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