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melisa1 [442]
3 years ago
15

8 out of every 20 spectators were girls. There were a total of 1560 spectators at the game. How many of the spectators were girl

s
Mathematics
2 answers:
kakasveta [241]3 years ago
8 0
The answer is 624 girls.
Oduvanchick [21]3 years ago
7 0

Answer: I think the answer is 624 girls im sorry if its wrong if it is can you please correct me

Step-by-step explanation:

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2.2a+3.4b=1.2a+2.6 solve for a
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A = 2.6-3.4b

Step-by-step explanation:

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Graph the area bounded by y =12, x>=0 and x<=12
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Answer:

Graph.

y=12

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x<=12

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There are 156 sixth graders and 7 sixth grade teachers.There are 120 fifth graders and 5 fifth grade teachers.Which grade has lo
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Fifth grade has 24 students per each teacher.
Sixth grade has about 23 students per teacher.

Sixth grade is lower.
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Ajar contains 17 blue cubes, 4 blue spheres, 5 green cubes, and 16 green spheres. What is the probability of randomly selecting
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Answer:

The answer would be 3/16

Step-by-step explanation:

Total no.of beads = 16  

No.of blue beads = 3  

Probability of picking a blue bead = 3/16

8 0
3 years ago
In a recent study on world​ happiness, participants were asked to evaluate their current lives on a scale from 0 to​ 10, where 0
uysha [10]

Answer:

a) A response of 8.9 represents the 92nd ​percentile.

b) A response of 6.6 represents the 62nd ​percentile.

c) A response of 4.4 represents the first ​quartile.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 5.9

Standard Deviation, σ = 2.2

We assume that the distribution of response is a bell shaped distribution that is a normal distribution.

Formula:

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

a) We have to find the value of x such that the probability is 0.92

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.92

Calculation the value from standard normal z table, we have,  

P(z

\displaystyle\frac{x - 5.9}{2.2} = 1.405\\x = 8.991 \approx 8.9

A response of 8.9 represents the 92nd ​percentile.

b) We have to find the value of x such that the probability is 0.62

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.62

Calculation the value from standard normal z table, we have,  

P(z

\displaystyle\frac{x - 5.9}{2.2} = 0.305\\x = 6.571 \approx 6.6

A response of 6.6 represents the 62nd ​percentile.

c) We have to find the value of x such that the probability is 0.25

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.25

Calculation the value from standard normal z table, we have,  

P(z

\displaystyle\frac{x - 5.9}{2.2} = -0.674\\x = 4.4172 \approx 4.4

A response of 4.4 represents the first ​quartile.

4 0
3 years ago
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