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frez [133]
3 years ago
11

school fair ticket costs $8 per adult and $1 per child. On a certain day, the total number of adults (a) and children (c) who we

nt to the fair was 30, and the total money collected was $100. Which of the following options represents the number of children and the number of adults who attended the fair that day, and the pair of equations that can be solved to find the numbers? (4 points)
Mathematics
1 answer:
AnnyKZ [126]3 years ago
6 0

Answer:

12 adults and 4 kids were there that the fair

Step-by-step explanation:

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Help What is the solution?
VladimirAG [237]

Answer:

The answer is B.) x=1, y=2

Step-by-step explanation:

2(1)+2=4

3(1)+3(2)=9

3 0
2 years ago
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At a candy store you can get 3 giant lollipops for $4.89. How much would it cost to buy 10 lollipops?
Bingel [31]

Answer:

$16.30

Step-by-step explanation:

So if you can buy 3 lollipops for $4.89, then you can by 1 for $1.63. Then 10 lollipops would be $16.30.  

4 0
2 years ago
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Can you please help me please i need real people not robots
Anni [7]
I think is A or b go try it am not 100% sure
8 0
2 years ago
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F(x) = x – 5x + 2 – 3x + 6<br> State how many complex and how many zeros the function have?
Artemon [7]
I hope that was helpful

6 0
3 years ago
The mean cost of a five pound bag of shrimp is 40 dollars with a standard deviation of 7 dollars. If a sample of 51 bags of shri
damaskus [11]

Answer:

P(|\bar{x}-40| > 0.6)=0.5404

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 40 dollars

Standard Deviation, σ = 7 dollars

Sample size,n = 51

We are given that the distribution of cost of shrimp is a bell shaped distribution that is a normal distribution.

Formula:

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

Standard error due to sampling =

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{7}{\sqrt{51}} = 0.9801

P(sample mean would differ by true mean by more than 0.6)

P(|\bar{x}-40| > 0.6)\\\Rightarrow = 1-P(39.4

P(39.4 \leq x \leq 40.6) \\\\= P(\displaystyle\frac{39.4 - 40}{0.9801} \leq z \leq \displaystyle\frac{40.6-40}{0.9801})\\\\ = P(-0.6121 \leq z \leq 0.6121)\\= P(z \leq 0.6121) - P(z < -0.6121)\\= 0.7298 - 0.2702 =0.4596

P(|\bar{x}|-40 > 0.6)\\= 1-0.4596=0.5404

0.5404 is the required probability.

6 0
2 years ago
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