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Mariana [72]
4 years ago
7

Someone please Help me ASAP! You would be a blessing!

Mathematics
1 answer:
ozzi4 years ago
3 0

11. Let the Number of Students filled in One Van be : V

Let the Number of Students filled in One Bus be : B

Given : The Senior Class of High School A rented and filled 8 Vans and 3 Buses with 150 students

⇒ 8V + 3B = 150 --------------- [1]

Given : The Senior Class of High School B rented and filled 10 Vans and 3 Buses with 162 students

⇒ 10V + 3B = 162 --------------- [2]

Subtracting Equation [1] from Equation [2], we get :

⇒ (10V + 3B) - (8V + 3B) = 162 - 150

⇒ 10V - 8V + 3B - 3B = 12

⇒ 2V = 12

⇒ V = 6

Substituting V = 6 in Equation [1], We get :

⇒ 8(6) + 3B = 150

⇒ 48 + 3B = 150

⇒ 3B = 150 - 48

⇒ 3B = 102

⇒ B = 34

⇒ Each Van can carry 6 Students

⇒ Each Bus can carry 34 students

12 . Let the Price of One Adult Ticket be : A

Let the Price of One Student Ticket be : S

Given : On the First Day, The School sold 6 Adult tickets and  7 student tickets for a Total of $122

⇒ 6A + 7S = 122 ----------- [1]

Given : On the Second Day, The School sold 6 Adult tickets and  6 student tickets for a Total of $114

⇒ 6A + 6S = 114 ----------- [2]

Subtracting Equation [2} from Equation [1], We get :

⇒ (6A + 7S) - (6A + 6S) = 122 - 114

⇒ 6A - 6A + 7S - 6S = 8

⇒ S = 8

Substituting S = 8 in Equation [1], We get :

⇒ 6A + 7(8) = 122

⇒ 6A + 56 = 122

⇒ 6A = 122 - 56

⇒ 6A = 66

⇒ A = 11

⇒ Price of One Adult Ticket = $11

⇒ Price of One Student Ticket = $8

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Answer:

<em>(a) A 99% confidence interval for the actual mean noise level in hospitals is </em><em>(44.02 db, 49.98 db)</em><em>. </em>

<em>(b) We can be 90% confident that the actual mean noise level in hospitals is </em><em>47 db</em><em> with a margin of error of </em><em>1.89 db</em><em>. </em>

<em>(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between </em><em>44.41 db and 49.59 db</em><em>. </em>

<em />

Step-by-step explanation:

<em>The problem is incomplete. The questions are:</em>

<em />

<em>(a) A 99% confidence interval for the actual mean noise level in hospitals is </em><em>(44.02 db, 49.98 db)</em><em>. </em>

For a 99% CI, the value of z is z=2.58

Then, the confidence interval for the mean is:

M-z\sigma/\sqrt{n}\leq\mu\leq M-z\sigma/\sqrt{n}\\\\47-2.58*10/\sqrt{75}  \leq\mu\leq47+2.58*10/\sqrt{75}\\\\47-2.98\leq\mu\leq47+2.98\\\\44.02\leq\mu\leq 49.98

<em>(b) We can be 90% confident that the actual mean noise level in hospitals is </em><em>47 db</em><em> with a margin of error of </em><em>1.89 db</em><em>. </em>

For a 90% CI, the value of z is z=1.64.

Then, we can calculate the margin of error as:

E=z*\sigma/\sqrt{n}=1.64*10/\sqrt{75}=1.89

<em>(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between </em><em>44.41 db and 49.59 db</em><em>. </em>

The 2% tails data corresponds, in the standard normal distirbution, to the values of z whose absolute value is higher than 2.33.

The values of db for these critical values are:

X_1=M+z_1*\sigma/\sqrt{n}=47+(-2.33)*10/\sqrt{81}=47-2.59=44.41\\\\\\ X_2=M+z_2*\sigma/\sqrt{n}=47+(2.33)*10/\sqrt{81}=47+2.59=49.59

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