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Temka [501]
3 years ago
11

In a certain city, 80 percent of the households have cable television and 60 percent of the households have videocassette record

ers. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:
Mathematics
2 answers:
nevsk [136]3 years ago
5 0

Answer: 60000 households have both cable television and videocassette recorders.

Step-by-step explanation:

Let x represent the number of households that have both cable television and videocassette recorders.

In the city, 80 percent of the households have cable television.

If there are 150,000 households in the city, then the number of households that have cable television would be

80/100 × 150000 = 120000

Also, 60 percent of the households have videocassette recorders. It means that the number of households that have videocassette recorders would be

60/100 × 150000 = 90000

The number that have only only cable television is 120000 - x

The number that have only only videocassette recorders is

90000 - x

Therefore,

120000 - x + 90000 - x + x = 150000

210000 - x = 150000

x = 210000 - 150000

x = 60000

alexandr1967 [171]3 years ago
3 0

Answer:

60,000

Step-by-step explanation:

Let the total people in the household be 100%

If 80 percent of the households have cable television, 60 percent of the households have videocassette recorders and the number of households that have both cable television and videocassette recorders is 'x' percent. The number of households having ONLY cable television will be 80-x while households having ONLY videocassette will be 60-x.

To get x, we will have;

80-x + 60-x + x = 100

140-x = 100

x = 140-100

x = 40

This shows that 40% of the households have both cable television and videocassette.

If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be 40% of 150,000

= 40/100 × 150,000

= 60,000

Number of households that owns both television and videocassette could be from 60,000.

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Any point, <span><span>(<span><span>x0</span>,<span>y0</span></span>)</span><span>(<span><span>x0</span>,<span>y0</span></span>)</span></span> on the parabola satisfies the definition of parabola, so there are two distances to calculate:

<span>Distance between the point on the parabola to the focusDistance between the point on the parabola to the directrix</span>

To find the equation of the parabola, equate these two expressions and solve for <span><span>y0</span><span>y0</span></span> .

Find the equation of the parabola in the example above.

Distance between the point <span><span>(<span><span>x0</span>,<span>y0</span></span>)</span><span>(<span><span>x0</span>,<span>y0</span></span>)</span></span> and <span><span>(<span>a,b</span>)</span><span>(<span>a,b</span>)</span></span> :

<span><span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span></span><span>‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾</span>√</span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span></span></span>

Distance between point <span><span>(<span><span>x0</span>,<span>y0</span></span>)</span><span>(<span><span>x0</span>,<span>y0</span></span>)</span></span> and the line <span><span>y=c</span><span>y=c</span></span> :

<span><span><span>∣∣</span><span><span>y0</span>−c</span><span>∣∣</span></span><span>| <span><span>y0</span>−c</span> |</span></span>

(Here, the distance between the point and horizontal line is difference of their <span>yy</span> -coordinates.)

Equate the two expressions.

<span><span><span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span></span><span>‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾</span>√</span>=<span><span>∣∣</span><span><span>y0</span>−c</span><span>∣∣</span></span></span><span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span></span>=<span>| <span><span>y0</span>−c</span> |</span></span></span>

Square both sides.

<span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span>=<span><span>(<span><span>y0</span>−c</span>)</span>2</span></span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span>=<span><span>(<span><span>y0</span>−c</span>)</span>2</span></span></span>

Expand the expression in <span><span>y0</span><span>y0</span></span> on both sides and simplify.

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This equation in <span><span>(<span><span>x0</span>,<span>y0</span></span>)</span><span>(<span><span>x0</span>,<span>y0</span></span>)</span></span> is true for all other values on the parabola and hence we can rewrite with <span><span>(<span>x,y</span>)</span><span>(<span>x,y</span>)</span></span> .

Therefore, the equation of the parabola with focus <span><span>(<span>a,b</span>)</span><span>(<span>a,b</span>)</span></span> and directrix <span><span>y=c</span><span>y=c</span></span> is

<span><span><span><span>(<span>x−a</span>)</span>2</span>+<span>b2</span>−<span>c2</span>=2<span>(<span>b−c</span>)</span>y</span></span>

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