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2 years ago

Someone can help me for all the question ?

Mathematics

1 answer:

Dvinal [7]2 years ago

4 0

Answer:

1. 5

2. 3 adult tickets; 2 child tickets

3. See below

Step-by-step explanation:

2 adult tickets cost 2 × $12 = $24

Subtract $24 from $64:

$64 - $24 = $40

Divide $40 by $8:

$40 ÷ $8 = 5

Answer: 5 child tickets

x = number of adult tickets

y = number of child tickets

The total number of tickets is 5, so the first equation is:

x + y = 5

The cost of x adult tickets is 12x. The cost of y child tickets is 8y. The total cost is 12x + 8y. We are told the total cost is $52. The second equation is:

12x + 8y = 52

The system of equation is:

x + y = 5

12x + 8y = 52

Let's solve it by substitution.

Solve the first equation for x.

x = 5 - y

Now we substitute x in the second equation by 5 - y.

12(5 - y) + 8y = 52

Distribute:

60 - 12y + 8y = 52

Combine like terms on the left side:

-4y + 60 = 52

Subtract 60 from both sides:

-4y = -8

Divide both sides by 4.

y = 2

Substitute y = 2 in the first original equation and solve for x.

x + y = 5

x + 2 = 5

x = 3

Answer: 3 adult tickets; 2 child tickets

The customer's claim is incorrect. His sum is $71. 71 is an odd number. Since all ticket prices are even numbers, it is impossible to add only even numbers and get an odd sum. A sum of several amounts of 12 and several amounts of 8 can never equal 71.

We can also show his claim is false using a system of equations.

x + y = 7

12x + 8y = 71

Use substitution.

x = 7 - y

12(7 - y) + 8y = 71

84 - 12y + 8y = 71

84 - 4y = 71

-4y = -13

y = 3.25

This means 3.25 child tickets. The problem states that tickets can be purchased only in whole numbers, so 3.25 tickets cannot be purchased, so the customer's claim is false.

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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

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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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saul85 [17]

Answer:

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