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Sophie [7]
3 years ago
5

At a local concert, the cost for 3 adults and 2 children was $32.00. The cost for 8 adults and 5 children

Mathematics
1 answer:
Nuetrik [128]3 years ago
8 0
The adult ticket price is $8.00 and the child ticket price is $4.00 :)
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Solve for c.<br><br> 3 − 7c − 20c = 7(–7c − 19) + 14c<br><br> c =
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Answer:

\boxed{c = -17}

Step-by-step explanation:

=  > 3 - 7c - 20c = 7( - 7c - 19) + 14c \\  \\  =  > 3 - 27c = ( - 7c \times 7) - (7 \times 19) + 14c \\  \\  =  > 3 - 27c =  - 49c - 133 + 14c \\  \\  =  > 3 - 27c =  - 35c - 133 \\  \\  =  > 3 - 27c + 35c =  - 133 \\  \\  =  > 3  + 8c =  - 133 \\  \\  =  > 8c =  - 133 - 3 \\  \\  =  > 8c =  - 136 \\  \\  =  > c =  -  \frac{136}{8}  \\  \\  =  > c =  - 17

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3 years ago
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You invest in a new play. The cost includes an overhead of $30,000, plus production costs of $2500 per performance. A sold-out p
Alchen [17]

Answer: 48 sold-out performances must be played in order for you to break even.

Step-by-step explanation:

Let x represent the number of sold-out performances must be played in order for you to break even.

You invest in a new play. The cost includes an overhead of $30,000, plus production costs of $2500 per performance. This means that the total cost of x sold out performances would be

2500x + 30000

A sold-out performance brings you $3125. This means that he total revenue from x sold out performances would be

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Therefore,

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3 years ago
The national mean annual salary for a school administrator is $90,00 a year (The Cincinnati Enquirer, April 7, 2012). A school o
timurjin [86]

Answer:

a) Null hypothesis:\mu = 90000  

Alternative hypothesis:\mu \neq 90000  

b) p_v =2*P(t_{(24)}  

c) If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

Step-by-step explanation:

1) Data given and notation  

77600 ,76000 ,90700 ,97200 ,90700 ,101800 ,78700 ,81300 ,84200 ,97600 ,

77500 ,75700 ,89400 ,84300 ,78700 ,84600 ,87700 ,103400 ,83800 ,101300

94700 ,69200 ,95400 ,61500 ,68800

We can calculate the sample mean and deviation with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

The values obtained are:

\bar X=85272 represent the mean annual salary for the sample  

s=11039.23 represent the sample standard deviation for the sample  

n=25 sample size  

\mu_o =90000 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean salary differs from 90000, the system of hypothesis would be:  

Null hypothesis:\mu = 90000  

Alternative hypothesis:\mu \neq 90000  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part b: Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{85272-90000}{\frac{11039.23}{\sqrt{25}}}=-2.141    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(24)}  

Part c: Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

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