Mathematics, 18.10.2019 23:30 kenisonpaigebosma

Suppose that 55% of all adults regularly consume coffee,
45% regularly consume carbonated soda, and 70% regularly
consume at least one of these two products.
a) what is the probability that a randomly selected adult
regularly consumes both coffee and soda?
b. what is the probability that a randomly selected adult
doesn’t regularly consume at least one of these two
products?

Answers: 1

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Answers

The probability that a randomly selected adult doesn’t regularly consume at least one of these two products is 30%

Be:
A: Adults regularly consume coffee
B: Adults regularly consume carbonated soda
C: Adults doesn't regularly consume at leas one of these two products.

55% of all adults regularly consume coffee→P(A)=55%=55/100→P(A)=0.55
45% regularly consume carbonated soda→P(B)=45%=45/100→P(B)=0.45
70% regularly consume at least one of these two products→P(A U B)=70%=70/100→P(A U B)=0.70

a) What is the probability that a randomly selected adult regularly consumes both coffee and soda?
P(A ∩ B)=?

P(A U B)=P(A) + P(B) - P(A ∩ B)

Replacing P(A U B)=0.70; P(A)=0.55; and P(B)=0.45 in the equation above:
0.70=0.55 + 0.45 - P(A ∩ B)→
0.70=1.00 - P(A ∩ B)

Solving for P(A ∩ B): Subtracting 1.00 both sides of the equation:
0.70-1.00=1.00 - P(A ∩ B) -1.00→
-0.30 = - P(A ∩ B)

Multiplying both sides of the equation by (-1):
(-1)(-0.30) = (-1)[ - P(A ∩ B)]→
0.30 = P(A ∩ B)→
P(A ∩ B) =0.30→
P(A ∩ B) = (0.30)*100%→
P(A ∩ B) = 30%

The probability that a randomly selected adult regularly consumes both coffee and soda is 70%

b. What is the probability that a randomly selected adult doesn’t regularly consume at least one of these two products?
P(C)=?

P(A U B) + P(C)=1

Replacing P(A U B)= 0.70 in the equation above:
0.70+P(C)=1

Solving for P(C). Subtracting 0.70 both sides of the equation:
0.70+P(C)-0.70=1-0.70→
P(C)=0.30→
P(C)=(0.30)*100%→
P(C)=30%

The probability that a randomly selected adult doesn’t regularly consume at least one of these two products is 30%

Answer from: Quest

yes the data was not prsented

step-by-step explanation:

Answer from: Quest

answer:

total cost for tiles and paints is $924.

step-by-step explanation:

we have been given that a community hall is in the shape of a cuboid. the hall is 40m long 15m high and 3m wide.

the paint will be required for 4 walls and ceiling.

let us find area of walls and ceiling.

$\text{area of walls and ceiling}=(2*40*15)+(2*3*15)+(40*3)$

$\text{area of walls and ceiling}=1200+90+120$

$\text{area of walls and ceiling}=1410$

therefore, the area of walls and ceiling is 1410 square meters.

given: cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. therefore,

$\text{ the total painting cost}=10*(\frac{1410}{25})$

$\text{ the total painting cost}=10*56.4=564$

therefore, the total painting cost is $564.

tiles will be required for floor. let us find the area of floor.

$\text{area of floor} = 40*3\text{ square meters}$

$\text{area of floor} =120\text{ square meters}$

given: 1m squared floor tiles costs $3. so,

$\text{total cost for tiles} = 3*120 = 360$

therefore, the total tiles cost is $360.

now let us find combined total cost of tiles and paint.

$\text{combined total cost}= 564+360 = 924$

therefore, the combined total cost of tiles and paint is $924.

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Suppose that 55% of all adults regularly consume coffee,
45% regularly consume carbonated soda...

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