Mathematics, 14.09.2019 07:30 mponcier4

a special-interest group has conducted a survey concerning a ban on hand guns. note: a rifle is a gun, but it is not a hand gun. the survey yielded the following results for the 1000 households that responded.
267 own a hand gun.
437 own a rifle.
493 supported the ban on hand guns.
136 own both a hand gun and a rifle.
206 own a rifle but no hand gun and do not support the ban on hand guns.
70 own a hand gun and support the ban on hand guns.
48 own both a hand gun and a rifle and also support the ban on hand guns.
how many of the surveyed households:
(a) only own a hand gun and do not support the ban on hand guns?
(b) do not own a gun and support the ban on hand guns?
(c) do not own a gun and do not support the ban on hand guns?

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Answer from: TrapQueen665

a)109 of the surveyed households only own a hand gun and do not support the ban on hand guns.

b)328 of the surveyed households do not own a gun and support the ban on hand guns

c )112 of the surveyed households do not own a gun and do not support the ban on handguns.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents those who own a hand gun.

-The set B represents those who own a rifle

-The set C represents those who support the ban of handguns.

-D represents those who do not own a gun and do not support the ban on hand guns.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which are those that only own a hand gun, $A \cap B$ are those who own both a handgun and a rifle, $A \cap C$ are those who own a hand gun and support the ban on handguns, and $A \cap B \cap C$ are those who own both a hand gun and a rifle, and also support the ban on hand guns.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

In which are those who only own a rifle, and $B \cap C$ are those who own a rifle and support the ban on handguns.

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

In which are those who support the ban on hand guns and do not own any handguns.

This diagram has the following subsets:

$a,b,c,D(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

There were 1000 households surveyed, so:

$a + b + c + D + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1000$

We start finding the values from the intersection of three sets.

Solution:

48 own both a hand gun and a rifle and also support the ban on hand guns.

$A \cap B \cap C = 48$

70 own a hand gun and support the ban on hand guns:

$(A \cap C) + (A \cap B \cap C) = 70$

$A \cap C = 70 - 48$

$A \cap C = 22$

206 own a rifle but no hand gun and do not support the ban on hand guns.

b = 206 .

136 own both a hand gun and a rifle:

$(A \cap B) + (A \cap B \cap C) = 136$

$A \cap B = 136 - 48$

$A \cap B = 88$

493 supported the ban on hand guns.

C = 493

437 own a rifle:

B = 437

267 own a hand gun:

A = 267

a) How many of the surveyed households only own a hand gun and do not support the ban on hand guns?

This is the value of , that we can find from the following equation:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

267 = a + 88 + 22 + 48

a = 267 - 158

a = 109

109 of the surveyed households only own a hand gun and do not support the ban on hand guns.

b) How many of the surveyed households do not own a gun and support the ban on hand guns?

This is the value of , that we can find from the following equation:

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$493 = c + 22 + (B \cap C) + 48$

Here, we also have to find $B \cap C$ , that are those who own a rifle and support the ban on hand guns. We can find this from the following equation:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$437 = 206 + (B \cap C) + 88 + 48$

$B \cap C = 437 - 342$

$B \cap C = 95$

$493 = c + 22 + (B \cap C) + 48$

493 = c + 22 + 95 + 48

c = 493 - 165

c = 328

328 of the surveyed households do not own a gun and support the ban on hand guns.

a) How many of the surveyed households do not own a gun and do not support the ban on hand guns?

This is the value of d, that can be found from the following equation:

$a + b + c + D + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1000$

109 + 206 + 328 + D + 88 + 22 + 95 + 40 = 1000

D = 1000 - 888

D = 112

112 of the surveyed households do not own a gun and do not support the ban on handguns.

Answer from: Quest

answer: i think one of the answer is $150 and $300?

step-by-step explanation:

Answer from: Quest

answer: 35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %, p(e) = 27 % and $p(l\cap e)$ = 4 %

and, we have to find out, $p(l\cup e)$

but, we know that,

$p(l\cup e)= p(l) + p(e) - p(l\cap e)$

which is the required equation that will use to find the probability of a late departure or an early departure.

$p(l\cup e)= 12 + 27 - 4$

$p(l\cup e)= 35 %$

therefore, the probability of a late departure or an early arrival = 35 %

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