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2 years ago

200 people are asked about three brands of beverages Coffee, Tea and Chocolate. 36 like

Mathematics

2 answers:

sergiy2304 [10]2 years ago

6 0

Answer:

i) 52

ii) 66

Step-by-step explanation:

This sort of problem is usually solved using some sort of table or Venn diagram. We have used a table in the attachment such that each of the cells in the grid represents one of the 8 possible combinations of coffee, tea, or chocolate.

The numbers in black are given directly by the problem statement. The numbers in blue are calculated from the numbers in black and one of the other given numbers in the problem statement. The numbers in red are calculated so the totals work out correctly.

The cells shaded blue-gray are "one and only one" of the drinks. The cell shaded purple-gray is "don't like any".

<h3>i) black coffee</h3>

We assume that any form of "coffee" is "black coffee." The problem statement tells us 52 people like coffee.

The last number we calculated is the number who like none of the drinks: 66. This brings the total of non-coffee drinkers to 148.

_____

<em>Additional comment</em>

As noted above, the colors more or less indicate the order of the calculations. We started with 36 who like coffee only, and 46 who like coffee and not tea. That means 46-36=10 like coffee and chocolate, but not tea.

Since 16 like coffee and chocolate, that leaves 16-10=6 who like coffee and tea and chocolate (all three).

96 like chocolate, but 16 of those like chocolate and tea. The other 80 like chocolate, but not tea.

46+16=52, so there must be 0 who like coffee and tea, but not chocolate.

The number who like only one is 108. We already have 36 that are coffee only, 80-10 = 70 that are chocolate only, so there must be 108 -70 -36 = 2 who like tea only.

Now, we can finish filling the top row of the table and the totals in the bottom row.

Scorpion4ik [409]2 years ago

4 0

Answer:

Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

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For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

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$y = m\cdot x + b$

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Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

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$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

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The equation of the line in the yz-plane is $z = -\frac{3}{2}\cdot y+3$ or $3\cdot y + 2\cdot z = 6$ .

xz-plane (2, 0, 0) and (0, 0, 3)

$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - z-Intercept, dimensionless.

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Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

x-Intercept

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$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the xz-plane is $z = -\frac{3}{2}\cdot x+3$ or $3\cdot x + 2\cdot z = 6$

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