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3 years ago

14

a popular movie debuted in 2940 theaters and earned an average of $12,490 per cinema the weekend of the debut. Find the amount t

hat the movie earned in its first weekend.

1 answer:

Yuri [45]3 years ago

8 0

Hello! So to find the amount the movie earned that first weekend, all you have to do is multiply the amount of theaters by the amount made per theater. In this case, there are 2,940 theaters and made an average of $12,490. 2,940 * 12,490 is 36,720,600. There. The movie made $36,720,600 on it's opening weekend.

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erica [24]

Answer:

I hope it helps it was the best I can do.

Step-by-step explanation:

-x12 + x7 = 0

Factorization:

−x7(x − 1)(x4 + x3 + x2 + x + 1) = 0

Solutions based on Jenkins–Traub algorithm:

x1 = … = x7 = 0

x8 = 1

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3 years ago

Help plz <br> How do you solve

tensa zangetsu [6.8K]

Answer:

$y=4\sqrt{3}\ units$

Step-by-step explanation:

Triangle MRN is created by folding an equilateral triangle by half. Since triangle MRN is right triangle, the equilateral triangle must be folded along the height drawn to one side.

In equilateral triangle with side length of a units, the height is $\dfrac{a\sqrt{3}}{2}\ units$ and the height is greater than half the side because

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The leg with length x units has the smaller projection (2 units long) than the leg with length y units has (6 units long).

This means the leg with length y units is the height of the equilateral triangle. If the leg with length y units is the height of the equilateral triangle, then the hypotenuse MN = 8 units is the side of the equilateral triangle and lex with length x units is half the side, so x = 4 units.

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3 years ago

Denise, Keith, and Tim live in the same neighborhood. Denise lives 0.3 mile from Keith. The distance that Tim and Keith live fro

hodyreva [135]

Answer:

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Step-by-step explanation:easy

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2 years ago

a survey showed that 82% of youth most often use the internet at home . what fraction of youth surveyed use the internet most of

Yuliya22 [10]

When talking about surveys (and most other things) we use 100% as the cap. You can not work harder than 100% efficiency and you can not have more than 100% of people agree with you. You can think of this as a pie.

A whole pie would be 100%
A half a pie 50%
Etc.

You can not have more than a whole pie (without adding another pie or variable in this case)

So when you look at this problem with that analogy in mind, all it is is simple addition and subtraction.

If you create a survey and 82% of people vote yes, how many, out of 100%, voted yes.

To find this we take 100% (The whole pie or everybody who voted) and - the portion of the pie we ate or everyone who voted yes, 82%, which will give us the answer, everyone who voted no or the pie that is left on the platter.

So your equation would be 100 - 82 =18

18% of people interview used internet someplace other than in the home. <span />

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3 years ago

A recent health report revealed that a woman with insurance spends an average of 2.3 days in the hospital following a routine ch

cupoosta [38]

Answer:

$t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385$

$p_v =2*P(t_{30}>2.385)=0.0236$

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

$(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611$

$(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861$

Step-by-step explanation:

Data given and notation

$\bar X_{insurance}=2.3$ represent the mean for insurance

$\bar X_{No. ins}=1.9$ represent the mean withour insurance

$s_{Insurance}=0.6$ represent the sample standard deviation for the insurance case

$s_{No. Ins}=0.3$ represent the sample standard deviation for the No insurance case

$n_{Insurance}=16$ sample size for insurance

$n_{No. Iss}=16$ sample size for no insurance

t would represent the statistic (variable of interest)

$\alpha=0.01$ 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 the two groups are different, the system of hypothesis would be:

Null hypothesis: $\mu_{Insu}=\mu_{No. Ins}$

Alternative hypothesis: $\mu_{Ins} \neq \mu_{No. Ins}$

Since we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

$t=\frac{\bar X_{Ins}-\bar X_{No.Ins}}{\sqrt{\frac{s^2_{Ins}}{n_{Ins}}+\frac{s^2_{No. Ins}}{n_{No. Ins}}}}$ (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.

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:

$t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385$

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

$df = n_{ins} +n_{No Ins}-2=16+16-2 =30$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{30}>2.385)=0.0236$

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

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

$(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611$

$(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861$

6 0

3 years ago