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cestrela7 [59]
3 years ago
8

A comedy show sells lower level tickets for $77 and upper level tickets for $99. On the opening night 2615 tickets were sold for

a total $247071.
Mathematics
2 answers:
babunello [35]3 years ago
7 0

Let x represent lower level tickets that cost $77

Let y represent upper level tickets that cost $99

Cost equation: 77x + 99y = 247,071

Tickets equation: x + y = 2615

Using the elimination method, multiply the second equation by -77:

77x + 99y = 247,071

-77x - 77y = -201,355

--> 22y = 45,716

--> y = 2,078

Now plug "y" into either equation and solve for "x". I chose the Tickets equation. x + y = 2615 → x = 2615 - y → x = 2615 - 2078 → x = 537

Answer: lower level = 537 tickets, upper level = 2078 tickets.


rusak2 [61]3 years ago
5 0

Let U = upper

Let L = lower

U + L = 2615

99U + 77L = 247071 You can make your life slightly easier by dividing this equation through by 11

9U + 7L =24461 Multiply the first equation by 7

7U + 7L = 2615 * 7

7U +7L = 18305

====================

Combine the 2 new equations

9U + 7L = 24461

<u>7U + 7L = 18305</u> Subtract

2U = 6156 Divide by 2

U = 6156 / 2

U = 3078

The Number of Upper Level Tickets sold was 3078

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Answer:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

S_p=2.940

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

df=60+72-2=130

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

n_1 =60 represent the sample size for people who used a self service

n_2 =72 represent the sample size for people who used a cashier

\bar X_1 =5.2 represent the sample mean for people who used a self service

\bar X_2 =6.1 represent the sample mean people who used a cashier

s_1=3.1 represent the sample standard deviation for people who used a self service

s_2=2.8 represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

And t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

System of hypothesis

Null hypothesis: \mu_1 \geq \mu_2

Alternative hypothesis: \mu_1 < \mu_2

This system is equivalent to:

Null hypothesis: \mu_1 - \mu_2 \geq 0

Alternative hypothesis: \mu_1 -\mu_2 < 0

We can find the pooled variance:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

And the deviation would be just the square root of the variance:

S_p=2.940

The statistic is given by:

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

The degrees of freedom are given by:

df=60+72-2=130

And now we can calculate the p value with:

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

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