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3 years ago

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A diner has collected data about customer coffee-drinking habits. They have calculated that P(cream) = 0.5, P(sugar) = 0.6, and

P(cream or sugar) = 0.7. Determine the P(cream and sugar). (2 points)

2 answers:

frozen [14]3 years ago

5 0

Answer:

P(cream and sugar) = 0.4

Step-by-step explanation:

* Lets study the meaning of or , and on probability

- The use of the word or means that you are calculating the probability

that either event A or event B happened

- Both events do not have to happen

- The use the word and, means that both event A and B have to happen

* The addition rules are:

# P(A or B) = P(A) + P(B) ⇒ mutually exclusive (events cannot happen

at the same time)

# P(A or B) = P(A) + P(B) - P(A and B) ⇒ non-mutually exclusive (if they

have at least one outcome in common)

- The union is written as A∪B or “A or B”.

- The intersection is written as A∩B or “A and B”

* Lets solve the question

∵ P(cream) = 0.5

∵ P(sugar) = 0.6

∵ P(cream or sugar) = 0.7

- To find P(cream and sugar) lets use the rule of non-mutually exclusive

∵ P(A or B) = P(A) + P(B) - P(A and B)

∴ P(cream or sugar) = P(cream) + P(sugar) - P(cream and sugar)

- Lets substitute the values of P(cream) , P(sugar) , P(cream or sugar)

in the rule

∵ 0.7 = 0.5 + 0.6 - P(cream and sugar) ⇒ add the like terms

∴ 0.7 = 1.1 - P(cream and sugar) ⇒ subtract 1.1 from both sides

∴ 0.7 - 1.1 = - P(cream and sugar)

∴ - 0.4 = - P(cream and sugar) ⇒ multiply both sides by -1

∴ 0.4 = P(cream and sugar)

* P(cream and sugar) = 0.4

EastWind [94]3 years ago

3 0

Answer:

0.4

Step-by-step explanation:

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The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

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First we need the equivalence between cups and pints to solve this question, the equivalence is:
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0.25 cup ---> x

x = (0.25)(0.5)/1
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Therefore 1/4 cups is equivalent to 0.125 pints

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