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3 years ago

PLEASE i AM CRYING RIGHT NOW I HAVE HAD NO SLEEP AND I DONT KNOW THE ANSWER PLEASE

Mathematics

1 answer:

MAVERICK [17]3 years ago

4 0

c = number of children tickets sold

a = number of adult tickets sold.

we know that 178 total tickets were sold, thus whatever "c" and "a" are, c + a = 178.

the price of a child ticket is $5.4, so the price for all "c" tickets must be 5.4c.

the price of a adult ticket is $9, so the price for all "a" tickets must be ac.

we also know the total sales for all tickets combined is $1310.4, so then 5.4c + 9a = 1310.4.

$\bf \begin{cases} c+a=178\implies \boxed{c}=178-c\\ 5.4c+9a=1310.4\\[-0.5em] \hrulefill\\ 5.4c+9\left( \boxed{178-c} \right)=1310.4 \end{cases} \\\\\\ 5.4c+1602-9c=1310.4\implies -3.6c+1602=1310.4\implies -3.6c=-291.6 \\\\\\ c=\cfrac{-291.6}{3.6}\implies c=81$

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Answer:

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Step-by-step explanation:

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Therefore, option C is correct from the given question.

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3 years ago

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Answer:

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

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If we compare the p value and the significance level provided we see that $p_v$ so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=35$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s =2.12$ represent the sample deviation obtained

$\sigma_0 =3$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is lower than 9 and the deviation lower than 3, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 9$

Alternative hypothesis: $\sigma^2$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case df= n-1= 35-1=34. And since is a left tailed test the p value would be given by:

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(16.979,34,TRUE)"

Conclusion

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3 years ago

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Step-by-step explanation:

<u>Step 1: Convert words into an expression</u>

52 minus 4 times the difference of 13 and 7.

(52 - 4) * (13 - 7)

Answer: (52 - 4) * (13 - 7)

If needed simplify

(52 - 4) * (13 - 7)

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Step-by-step explanation:

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