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rewona [7]
2 years ago
9

If 55% people in a city like Cricket, 30% like Football and the remaining like nother games, then what per cent of the people li

ke other games? If the total people is 60 lakh, find the exact number who like each type of games Answer it fast, no spam please
write in notebook ​
Mathematics
1 answer:
labwork [276]2 years ago
6 0

\large{\dag\:{\underline{\underline{\frak{\pmb{\red{Answer:-}}}}}}}

  • Cricket = 55%
  • Football = 30%
  • Other games = Remaining (?)

So, the percent of people who like <u>other games</u> equals:

= 100 – (55 + 30)

= 100 – 85

= 15%

<u>If </u><u>the </u><u>total </u><u>no.</u><u> </u><u>of.</u><u> </u><u>people</u><u> </u><u>is </u><u>6</u><u>0</u><u>,</u><u>0</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u>:

★ Cricket

= 60,00,000 × 55/100

= 33,00,000

★ Football

= 60,00,000 × 30/100

= 18,00,000

★ Other sports

= 60,00,000 × 15/100

= 9,00,000

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BlackZzzverrR [31]

Answer:

(1) The value of P (A) is 0.4286.

(2) The value of P (B) is 0.50.

(3) The value of P (A ∩ B) is 0.2143.

(4) The the value of P (B|A) is 0.50.

(5) The events <em>A</em> and <em>B</em> are independent.

Step-by-step explanation:

The events are defined as follows:

<em>A</em> = a student in the class has a sister

<em>B</em> = a student has a brother

The information provided is:

<em>N</em> = 210

n (A) = 90

n (B) = 105

n (A ∩ B) = 45

The probability of an event <em>E</em> is the ratio of the favorable number of outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

The conditional probability of an event <em>X</em> provided that another event <em>Y</em> has already occurred is:

P(X|Y)=\frac{P(A\cap Y)}{P(Y)}

If the events <em>X</em> and <em>Y</em> are independent then,

P(X|Y)=P(X)

(1)

Compute the probability of event <em>A</em> as follows:

P(A)=\frac{n(A)}{N}\\\\=\frac{90}{210}\\\\=0.4286

The value of P (A) is 0.4286.

(2)

Compute the probability of event <em>B</em> as follows:

P(B)=\frac{n(B)}{N}\\\\=\frac{105}{210}\\\\=0.50

The value of P (B) is 0.50.

(3)

Compute the probability of event <em>A</em> and <em>B</em> as follows:

P(A\cap B)=\frac{n(A\cap B)}{N}\\\\=\frac{45}{210}\\\\=0.2143

The value of P (A ∩ B) is 0.2143.

(4)

Compute the probability of <em>B</em> given <em>A</em> as follows:

P(B|A)=\frac{P(A\cap B)}{P(A)}\\\\=\frac{0.2143}{0.4286}\\\\=0.50

The the value of P (B|A) is 0.50.

(5)

The value of P (B|A) = 0.50 = P (B).

Thus, the events <em>A</em> and <em>B</em> are independent.

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Answer:

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Step-by-step explanation:

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2 years ago
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Pls mark me brainiless


(4/5)(x - 2) = 18
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3 years ago
Three nurses attended a seminar for two days. The travel expenses for each nurse were $982. Additional expenses per nurse per da
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3 years ago
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natta225 [31]

Answer:

A (I believe)

Step-by-step explanation:

since the part with the triangle is 6 yards, and it looks about half of that whole side, I'm going to assume that that side is 12 yards, and since it's a square each side will be the same length. so multiply 12 and 12 to get 144.

Now divide that by four so you can get the area of a fourth of the square and find the area of the triangle. 144 ÷ 4 = 36

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