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Sav [38]
2 years ago
7

A theater holds 2,680 seats. The lower level has 100 less than 2 times as many seats as the upper level. The middle level has 80

more than 3 times as many seats as the upper level. How many seats are in the middle level? 450 seats 800 seats 1,250 seats 1,430 seatsA theater holds 2,680 seats. The lower level has 100 less than 2 times as many seats as the upper level. The middle level has 80 more than 3 times as many seats as the upper level. How many seats are in the middle level?
A 450 seats
B 800 seats
C 1,250 seats
D 1,430 seats
Mathematics
2 answers:
Dimas [21]2 years ago
7 0

Answer:

Hello!!! Princess Sakura here ^^

Step-by-step explanation:

If the lower level has 100 less than 2 times as many seats as the upper level then the expression is 2x-100 and the middle level has 80 more than 3 times as many seats as the upper level so it's 3x+80 and lastly the top level is just x.

so now for the equation you would add all those expressions and they all should equal 2680:

x+3x+80+2x-100=2680\\6x-20=2680\\6x=2700\\x=450

but now don't forget to plug in for the middle level:

3x+80\\3(450)+80\\1350+80\\1430

Therefore the middle level has 1430 seats.

GaryK [48]2 years ago
5 0

Answer:

D is your answer fella

Step-by-step explanation:

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The mean weight of newborn infants at a community hospital is 6.6 pounds. A sample of seven infants is randomly selected and the
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Answer:

The correct option is  A

Step-by-step explanation:

From the question we are told that

    The  population is  \mu  =  6.6

     The level of significance is \alpha  =  5\%  = 0.05

      The sample data is  9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds

The Null hypothesis is H_o  :  \mu =  6.6

 The Alternative hypothesis is  H_a :  \mu  > 6.6

The critical value of the level of significance obtained from the normal distribution table is

                       Z_{\alpha } = Z_{0.05 } = 1.645

Generally the sample mean is mathematically evaluated as

      \=x =  \frac{\sum x_i }{n}

substituting values

      \=x =  \frac{9.0 +  7.3 +  6.0+ 8.8+ 6.8+ 8.4+6.6 }{7}

      \=x =  7.5571

The standard deviation is mathematically evaluated as

           \sigma  =  \sqrt{\frac{\sum  [ x -  \= x ]}{n} }

substituting values

          \sigma  =  \sqrt{\frac{  [ 9.0-7.5571]^2 + [7.3 -7.5571]^2 + [6.0-7.5571]^2 + [8.8- 7.5571]^2 + [6.8- 7.5571]^2 + [8.4 - 7.5571]^2+ [6.6- 7.5571]^2 }{7} }\sigma =  1.1774

Generally the test statistic is mathematically evaluated as

            t  =  \frac{\= x - \mu }  { \frac{\sigma }{\sqrt{n} } }

substituting values

           t  =  \frac{7.5571  - 6.6  }  { \frac{1.1774 }{\sqrt{7} } }

            t  = 1.4274

Looking at the value of  t and  Z_{\alpha }   we see that t  <  Z_{\alpha } hence we fail to reject the null hypothesis

  What this implies is that there is no sufficient evidence to state that the sample data show as significant increase in the average birth rate

The conclusion is that the mean is  \mu = 6.6  \ lb

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Answer:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

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And the confidence interval for the difference is between:

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Step-by-step explanation:

We have the following info given:

\bar X_1 = 240 sample mean for medium Pizzas from Prim's

\bar X_2 = 210 sample mean for medium Pizzas from Pizza Place

s_1 =8.6 sample deviation for Prim's

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n_1 =n_2 = 100  sample size selected for each case

The confidence interval for the difference of means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

And for the 95% confidence we need a significance level of \alpha=1-0.95=0.05 and \alpha/2 =0.025, the degrees of freedom are given by:

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And replacing we got:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047

And the confidence interval for the difference is between:

27.953 \leq \mu_1 -\mu_2 \leq 32.047

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​I'm not sure for part 3.
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2 years ago
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