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Goryan [66]
1 year ago
6

mr. larsen's third grade class has 22 students, 12 girls and 10 boys. two students must be selected at random to be in the fall

play. what is the probability that no boys will be chosen? order is not important.
Mathematics
1 answer:
faust18 [17]1 year ago
5 0

The probability that no boys will be chosen is 0.2857

What is probability?

The area of arithmetic called likelihood deals with numerical representations of the chance that an incident can occur or that a press release is true.

Main Body:

Total students =22

total boys = 10

total girls =12

Now we have to chose 2 girls from 12 , so combination is used to select ,

⇒¹²C₂

Also, We have to chose 2 girls from 22 students

so it can be represented as ,

⇒²²C₂

Probability of choosing 2 girls = ¹²C₂/²²C₂

On solving this we will get = 0.2857

Hence the probability is 0.2857

To know more about probability , visit;

brainly.com/question/13604758

#SPJ4

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Solve for x and y <br><br><br> This is a quiz please help
torisob [31]

Answer:

Hello! answer: y = 40 x = 67

Step-by-step explanation:

63 - 23 = 40 therefore y = 40

63 + 63 = 126 360 - 126 = 234 234 ÷ 2 = 117

67 × 2 = 134 134 - 17 = 117 therefore x = 67 hope that helps!

5 0
3 years ago
The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day
Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of  length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(between 236 and 281 days)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%

b) a) P(last between 236 and 296)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least 1-\dfrac{1}{k^2}  data lies within k standard deviation of mean.

For k = 2

1-\dfrac{1}{(2)^2} = 75\%

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

7 0
3 years ago
Tevon,leah,and beth are playing a computer game. Trevon scored -8. Beth's score was 3/4 of trever's score, and leah's score was
faust18 [17]

Answer:

Trevon scored -8.

Beth's score was 3/4 of trever's score = -8 *3/4 = -6

leah's score was 1/4 of beth's score. = -6 *1/4 = -1.5

Leah's score was -1.5


8 0
3 years ago
Simplify this expression.<br> 17 - 12x - 18 - 5x
sergij07 [2.7K]

Answer:

-17x-1

Step-by-step explanation:

5 0
3 years ago
The score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.
Artemon [7]

Answer:

P(X \geq 74) = 0.3707

Step-by-step explanation:

We are given that the score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Let X = Score of golfers

So, X ~ N(\mu=73,\sigma^{2}=3^{2})

The z score probability distribution is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = population mean = 73

           \sigma = standard deviation = 3

So, the probability that the score of golfer is at least 74 is given by = P(X \geq 74)

 P(X \geq 74) = P( \frac{X-\mu}{\sigma} \geq \frac{74-73}{3} ) = P(Z \geq 0.33) = 1 - P(Z < 0.33)

                                               =  1 - 0.62930 = 0.3707                  

Therefore, the probability that the score of golfer is at least 74 is 0.3707 .

3 0
3 years ago
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