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densk [106]
2 years ago
6

In a school annual Day 60 students involved is Science Exhibition, 75 involved in Mathematics Exhibition, 20 involved in both Ex

hibitions, and the rest of 285 students are not involved in any exhibition. Illustrate the above information in a Venn- diagrom. How many students took part in Schools Annual Day? How many of them involved in only one exhibition ?​
Mathematics
2 answers:
just olya [345]2 years ago
5 0

Answer:

1) See the explanation

2) 400 Students

3) 95 Students

Step-by-step explanation:

1) As can be known from the question

Then, draw the venn diagram according to the known conditions.

2) 60 + 75 - 20 = 115 { Draw a venn diagram to calculate }

115 + 285 = 400

3) 60 - 20 + 75 - 20 = 95 ( Draw a venn diagram to calculate }

pshichka [43]2 years ago
3 0

Answer:

440 man its so easy just add up all the numbers

Step-by-step explanation:

440

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This year, Shelley paid $2,363 for season tickets to her favorite baseball team, which is 15%
algol13

Answer:

I got 2,717.45

Step-by-step explanation:

I did 2,363 × 15% = 354.45

then 2,363 + 354.45 = 2,717.45

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3 years ago
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A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20
irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that p = 0.7

20 adult women

This means that n = 20

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

P(X < 14) = 1 - P(X \geq 14)

In which

P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916

P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789

P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304

P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716

P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278

P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068

P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008

So

P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079

P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

From the values found in item (a).

P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

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sorry I don't know

Step-by-step explanation:

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