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defon
3 years ago
12

Jon is selling tickets for the school talent show. On the 1st day, he sold 3 senior tickets and 12 child tickets for $195. On th

e 2nd day he sold 13 senior tickets for $299. Find the price of a senior citizen ticket.
Mathematics
1 answer:
Allisa [31]3 years ago
6 0
<h3>Question:</h3>

<em>Jon is selling tickets for the school talent show. On the 1st day, he sold 3 senior tickets and 12 child tickets for $195. On the 2nd day he sold 13 senior tickets for $299. Find the price of a senior citizen ticket.</em>

<h3>Answer:</h3>
  • $23

Create a system of equations to help you solve this problem. The system of equations will look like: 3s + 12c = 195 and 13s = 299. The variable s represents the cost of senior tickets and the variable c represents the cost of children tickets.

\left \{ {{3s~+~12c~=~195} \atop {13s~=~299}} \right.

Solve the second equation for the variable s as this is the easiest way to solve the problem. Solve the second equation for s by dividing both sides of the equation by 13 to isolate the variable s.

s = 23

Since the question was only asking for the price of a senior citizen ticket, you are technically done. The first equation was only put there to confuse you or allow you to check your work if you needed to. The price of a senior citizen ticket (variable s) is $23.

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At 8:00 am, the temperature was -10°F. The temperature dropped 3 degrees per hour for the next 5 hours. What was the temperature
sp2606 [1]

Answer:

-25 °F

Step-by-step explanation:

First, we would be multiply 3 by 5 to find the number we would be subtracting from the original

3 * 5 would give us 15

Then we subtract -10 by 15 and get -25 °F

The temperature at 1:00 PM would be -25 °F

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3 years ago
The following table summarises the data from a survey on the ownership of iPods among families with different levels of income.
AURORKA [14]

The two nominal variables are related.

The following table summarises the data from a survey on the ownership of iPods among families with different levels of income.  

Ownership           C1                  C2                  C3

No                         40                 32                   48

Yes                        30                 48                   52

The first thing to do in order to determine if they are related or not is to state our null and alternative hypothesis

Null hypothesis

\mathbf{H_o: Two \nomial  \ variables  \ are \  related}

Alternative hypothesis

\mathbf{H_a: Two \nomial  \ variables  \ are \  not \  related}

Using the Chi-square test statistics which can be expressed by using the formula  

X ^2 = \sum \dfrac{(O-E)^2}{E}

Ownership           C1                  C2                  C3                  Total

No                         40                 32                   48                 120

Yes                        30                 48                   52                 130

Total                      70                 80                   100                250

The expected values are calculated as:

\mathsf{E_{a,b} = \dfrac{(row \ total \times column \ total )}{grand \ total }}

\mathsf{E_{1,1} = \dfrac{(70 \times120 )}{250 }}

\mathsf{E_{1,1} = 33.6}

\mathsf{E_{1,2} = \dfrac{(70 \times130 )}{250 }}

\mathsf{E_{1,2} = 36.4}

\mathsf{E_{2,1} = \dfrac{(80 \times120 )}{250 }}

\mathsf{E_{2,1} = 38.4}

\mathsf{E_{2,2} = \dfrac{(80 \times 130 )}{250 }}

\mathsf{E_{2,2} = 41.6}

\mathsf{E_{3,1} = \dfrac{(100 \times120 )}{250 }}

\mathsf{E_{3,1} = 48}

\mathsf{E_{3,2} = \dfrac{(70 \times130 )}{250 }}

\mathsf{E_{3,2} = 52}

∴ Using the Chi-square test statistics, we have:

X ^2 = \sum \dfrac{(O-E)^2}{E}

X ^2 = \Bigg(  \dfrac{(40-33.6)^2}{33.6}+   \dfrac{(30-36.4)^2}{36.4}+   \dfrac{(32-38.4)^2}{38.4}+  \dfrac{(48-41.6)^2}{41.6} +   \dfrac{(48-48)^2}{48}+   \dfrac{(52-52)^2}{52} \Bigg)

X ^2 = \Bigg(  \dfrac{40.96}{33.6}+   \dfrac{40.96}{36.4}+   \dfrac{40.96}{38.4}+  \dfrac{40.96}{41.6} +   \dfrac{0}{48}+   \dfrac{0}{52} \Bigg)

X ^2 = \Bigg(  1.2190+   1.1253+  1.0667+ 0.9846+0+ 0 \Bigg)

\mathbf{X ^2 =4.3956}

The degree of freedom df = ((r - 1) × (c - 1))

= (3 - 1) (2 -1 )

= 2 × 1

= 2

∴

Assuming the level of significance = 5%

The p-value of the Chi-square test statistics at df of 2 is:

= \mathbf{P(X^2 > 4.3956) \implies 0.111}

Therefore, we can conclude that since the p-value (0.111) is greater than the level of significance (0.05), we fail to reject the null hypothesis.

Hence, the two nominal variables are related.

Learn more about Chi-square test statistics here:

brainly.com/question/2365682?referrer=searchResults

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3 years ago
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Answer:

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