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Nana76 [90]
3 years ago
6

Twelve students have equally likely chance of being selected in a drawing at school. Eight are girls and four are boys. What is

the probability that a boy will be selected?
Mathematics
2 answers:
ladessa [460]3 years ago
8 0
1/3 is the answer. 4/12 and that simplifies to 1/3
lubasha [3.4K]3 years ago
4 0
There are 4 boys and 12 students in total. It would be 4/12, but if it was to be simplified, it would be 1/3. The probability of a boy being selected is 1/3.
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A new car is purchased for $47, 000 and over time its value depreciates by one half
andrey2020 [161]

Answer:

$1500

Step-by-step explanation:

you can figure it out by going in increments of 4.5 years because that is the info you have from the problem so

47000 4.5 years later is 23500

9 years = 11750

13.5y = 5875

18y = 2937.5

22.5 years later is 1468.75

22.5 years is about 23 years and to the nearest hundred dollars that is $1500 so that is the answer

Hope this helps! :)

5 0
2 years ago
Read 2 more answers
The yearbook club is handing out T-shirts to its members. There are 5 blue, 7 green, 9 red, and 4 yellow T-shirts in all. If Jac
Neporo4naja [7]

Answer:

well it's 9/25 or 0.36% or (36%) only

Step-by-step explanation:

So you add all the numbers up it would be 25, divide the numerator by the denominator and get 0.36%, i can't remember if you divide .36 by 100 or not but anyway you get 36%.

But I hope i have helped you in anyway.

5 0
3 years ago
A bag contains 4 blue marbles and 2 yellow marbles. Two marbles are randomly chosen (the first marble is NOT replaced before dra
VMariaS [17]

Answer:

0.4 = 40% probability that both marbles are blue.

0.0667 = 6.67% probability that both marbles are yellow.

53.33% probability of one blue and then one yellow

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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)!}

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

What is the probability that both marbles are blue?

4 + 2 = 6 total marbles, which means that N = 6

4 blue, which means that k = 4

Sample of 2, which means that n = 2

This is P(X = 2). So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,6,2,4) = \frac{C_{4,2}*C_{2,0}}{C_{6,2}} = 0.4

0.4 = 40% probability that both marbles are blue

What is the probability that both marbles are yellow?

4 + 2 = 6 total marbles, which means that N = 6

2 yellow, which means that k = 2

Sample of 2, which means that n = 2

This is P(X = 2). So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,6,2,2) = \frac{C_{2,2}*C_{4,0}}{C_{6,2}} = 0.0667

0.0667 = 6.67% probability that both marbles are yellow.

What is the probability of one blue and then one yellow?

Total is 100%.

Can be:

Both blue(40%)

Both yellow(6.67%)

One blue and one yellow(x%). So

40 + 6.67 + x = 100

x = 100 - 46.67

x = 53.33

53.33% probability of one blue and then one yellow.

If you are told that both selected marbles are the same color, what is the probability that both are blue?

Conditional probability.

Event A: Both same color

Event B: Both blue

Probability of both being same color:

Both blue(40%)

Both yellow(6.67%)

This means that P(A) = 0.4 + 0.0667 = 0.4667

Probability of both being the same color and blue:

40% probability that both are blue, which means that P(A \cap B) = 0.4

Desired probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.4667} = 0.8571

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

8 0
2 years ago
0=9x^2-49<br> can someone show me how to find the zeroes step by step?
Molodets [167]

Answer: look at the picture

Step-by-step explanation: Hope this help :D

3 0
2 years ago
(2x^2 -x+5)v+ (3x^2 -3x -2)
babymother [125]
5x^2 - 4x +3
explanation is attached 
Enjoy !

7 0
2 years ago
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