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3 years ago

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If the average ticket price for movie A in its first week was 1.4 times the average ticket price in its third week, what was the

average ticket price for movie A in its first week?

2 answers:

LiRa [457]3 years ago

5 0

I think the answer will be 8.35

anyanavicka [17]3 years ago

3 0

Hello!

The correct answer is:

The average ticket price for movie A in its first week was $11.69.

<u><em> I hope you have an amazing day! :)</em></u>

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Step-by-step explanation:

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2 years ago

7 + (4 + 6) = 7+n name the propeerty

snow_tiger [21]

Answer:

n=15

Step-by-step explanation:

7+(4+6)=7+n

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3 years ago

What is 9x+12y<br> ._. <br> so can you help me ?? ...umm :p

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Simplifying

9x = 12y

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3 years ago

X + y + z = 7<br> 2x − y + z = 7<br> 3x − z = 2

soldi70 [24.7K]

<span>A) X + y + z = 7
B) 2x − y + z = 7

C) 3x − z = 2
</span><span>C) z = 3x -2 Substituting this into equations A) and B)

</span>
<span>A) X + y + 3x -2 = 7
B) 2x − y + 3x -2 = 7 Adding A and B
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</span>Since x = 2 then
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3 years ago

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances

Maurinko [17]

Answer:

$t=\frac{62-66}{\sqrt{\frac{10.5^2}{86}+\frac{12^2}{86}}}}=-2.326$

$p_v =P(t_{170}$

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, so the mean for the compound 1 is significantly less than the mean for the compound 2.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=62$ represent the mean for the compound 1

$\bar X_{2}=66$ represent the mean for the compound 2

$s_{1}=10.5$ represent the sample standard deviation for compound 1

$s_{2}=12$ represent the sample standard deviation for compund 2

$n_{1}=86$ sample size for the group compound 1

$n_{2}=86$ sample size for the group compound 2

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for compound 1 is less than the mean for compound 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}\geq\mu_{2}$

Alternative hypothesis: $\mu_{1} < \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value(s).

Based on the significance level $\alpha=0.05$ , we can find the critical values with the t distribution with 86+86-2=170 degrees of freedom, we are looking for values that accumulates 0.05 of the left tail of the distirbution.

For this case the value is $t_{\alpha}=-1.653$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{62-66}{\sqrt{\frac{10.5^2}{86}+\frac{12^2}{86}}}}=-2.326$

What is the p-value for this hypothesis test?

Since is a left tailed test the p value would be:

$p_v =P(t_{170}$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, so the mean for the compound 1 is significantly less than the mean for the compound 2.

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3 years ago