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3 years ago

A theater has 28 rows of 38 seats downstairs and 14 rows of 26 seats upstairs.How many seats does the theater have?

1 answer:

3 0

For downstairs- 28 x 38= 1064

For upstairs- 26 x 14= 364

All together- 1064 + 364= 1428

So.... 1428 seats in total.

Hope this helped! :)

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Find x Plz help me !!!

mel-nik [20]

Answer:

$\Huge\boxed{x=33.2}$

Step-by-step explanation:

Hello there!

We can solve for x using law of sines

As we can see in the image a side length divided by sin ( its opposite angle) = a different side length divided by sin ( its opposite angle)

So we can use this equation to solve for x

$\frac{21}{sin(35)} =\frac{x}{sin(65)}$

Our objective is to isolate the variable using inverse operations so to get rid of sin (65) we multiply each side by sin (65)

$\frac{x}{sin(65)} sin(65)=x\\\\\frac{21}{sin(35)} sin(65)=\frac{21sin(65)}{sin(35)}$

we're left with

$x=\frac{21sin(65)}{sin(35)}\\x=33.18208755$

assuming we have to round the answer would be 33.18 or 33.2

7 0

3 years ago

The list of all factors for 21 is 1, 3, 7, and 21. True or False plz i need help

antiseptic1488 [7]

Answer:

true

Step-by-step explanation:

hope you have a wonderful new year

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3 years ago

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Fynjy0 [20]

2/3 +4/3=6/3=2.... Self explanatory

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3 years ago

A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after

lana [24]

Answer:

95% confidence interval estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

(a) Lower Limit = 0.486

(b) Upper Limit = 0.624

Step-by-step explanation:

We are given that a statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's.

She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of persons who were still employed primarily as engineers = $\frac{111}{200}$ = 0.555

n = sample of graduates = 200

p = population proportion of engineers

Here for constructing 95% confidence interval we have used One-sample z proportion test statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.555-1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ , $0.555+1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ ]

= [0.486 , 0.624]

Therefore, 95% confidence interval for the estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

7 0

3 years ago