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Allushta
2 months ago
15

Suppose we pick three people at random. For each of the 2.32 The following questions, ignore the special case where someone migh

t be born on February 29th, and assume that births are evenly distributed throughout the year.(a) What is the probability that the first two people share a birthday?(b) What is the probability that at least two people share a birthday?
Mathematics
1 answer:
Inessa [12.5K]2 months ago
7 0

Answer:

(a) 1 in 365 or 0.2740%

(b) 0.8227%

Step-by-step explanation:

(a) For the first person's birthday, the probability that the second person has the same birthday is 1 out of 365, so the chance that the first two share a birthday is:

P = \frac{1}{365}=0.2740\%

(b) There are four scenarios possible where at least two individuals share a birthday: first and second, first and third, second and third, all three sharing the same birthday. Hence, the probability that at least two share their birthdays is:

P =3* \frac{1}{365}+ (\frac{1}{365})^2\\ P=0.8227\%

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In a school of 2100 students, the ratio of teachers to students is 1:14. Some teachers join the school and the ratio changes to
zzz [12365]

Answer:

50 Educators

Step-by-step explanation:

To tackle this question, the initial step is to calculate the amount of teachers prior to the addition of new staff. For this, I devised Model 1. In this model, teachers are positioned at the top of the ratio and students at the bottom. The variable X represents the number of teachers we are determining. Utilizing this model, I computed 2,100 multiplied by 1 (2,100) and then divided by 14 to conclude there were 150 teachers. Next, I formed a similar model with the updated student-teacher ratio (Model 2). This time, I multiplied 2,100 by 2 (which is 4,200) and divided by 21 to ascertain there are 200 teachers. Having established both the initial and the increased counts of educators, subtracting the original from the new gives you the tally of new teachers, which results in an increase of 50 teachers.

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1 month ago
Sara’s little brother is getting ready to color the picture of a rainbow that Sara drew for him. To make the task easier for her
Zina [12379]
There are seven rainbow colors: red, orange, yellow, green, blue, indigo, and violet, so 7 possible choices. When two events occur in sequence, multiply their probabilities. With replacement: P(violet)=1/7 and P(orange)=1/7, so P(violet then orange)=1/7 * 1/7 = 1/49. Without replacement: after picking violet, P(orange)=1/6, so P(violet then orange)=1/7 * 1/6 = 1/42.
8 0
18 days ago
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Simplify the given expression, using only positive exponents. Then complete the statements that follow. [ (x2y3)−1 (x−2y2z)2 ] 2
AnnZ [12381]

Answer: y²z⁴ / x¹²

  • The exponent for x is 12

  • The exponent for y is 2

  • The exponent for z is 4

Explanation:

1) The expression to simplify:

((x^2y^3)^{-1}(x^{-2}y^2z)^2)^2

2) Rule: power of a power:

(x^{-4}y^{-6})(x^{-8}y^8z^4)

3) Rule: product of powers with identical bases

x^{(-2-8)}y^{(-6+8)}z^4

4) Add up the exponents

x^{-12}y^2z^4

5) Transfer the negative exponent to the denominator:

\frac{y^2z^4}{x^{12}}

The exponent attached to x is 12, for y it is 2, and for z it is 4.

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1 month ago
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In the xy-plane, point E has coordinates (4,5) and point O has coordinates (0,0). Which of the following is an equation of the l
lawyer [12517]

Answer:

D

Step-by-step explanation:

You should insert the provided x and y coordinates into the equations to determine which one satisfies both points.

D is valid for both:

5 = 5/4(4)

0 = 5/4(0)

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2 months ago
Which steps can be used to solve StartFraction 6 Over 7 EndFraction x + one-half = StartFraction 7 Over 8 EndFraction for x? Che
lawyer [12517]

Response:

First, subtract one-half from each side of the equation.

Next, divide both sides by 6/7.

Then, multiply each side by 7/6.

Detailed explanation:

I simply took the question on ed.

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1 month ago
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