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kolezko [41]
4 years ago
15

The Jurassic Zoo charges ​$14 for each adult admission and ​$9 for each child. The total bill for the 214 people from a school t

rip was ​$2081. How many adults and how many children went to the​ zoo?  
Mathematics
1 answer:
Butoxors [25]4 years ago
4 0

a=adult

c=child

a+c=214

c=214-a

9c+14a=2081

9(214-a)+14a=2081

1926-9a+14a=2081

5a=155

a=155/5=31

31 adults

183 children


check

31*14 = 434

183*9=1647

1647+434=2081

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In a school, 30% of the students arrived by private car, 15 travel by taxi and the remainder walk to school. If the school has a
Yuliya22 [10]

Answer:

57.5 students

Step-by-step explanation:

Given that

The 30% students would be arrived by private car

So 15% would be travelled by taxi

And, the remaining i.e. 55% would be walk to school

Since the population of the school is 230

So by car, the students would be

= 230 × 0.30

= 69

By walk, the students would be

= 230 × 0.55  

= 126.50

Now the more students would be

= 126.50 - 69

= 57.5 students

6 0
3 years ago
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Sedbober [7]
The answer would be 6. 21/3.5=6
6 0
3 years ago
Read 2 more answers
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5 0
3 years ago
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puteri [66]

Answer:

2x+5

Step-by-step explanation:

The answer is in the question.

2 and x is 2x and plus 5 is +5.

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3 years ago
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A public health official is planning for the supplyof influenza vaccine needed for the upcoming flu season. She wants to estimat
Marizza181 [45]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25  

And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. And the critical value would be given by:

z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.04 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

We assume that a prior estimation for p would be \hat p =0.5 since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25  

And rounded up we have that n=421

5 0
4 years ago
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