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Setler79 [48]
3 years ago
15

A theatre has seats arranged in rows of 24, three hundred eighty -two(382) people attend a show at the theatre. What is the smal

lest number of rows that is needed to seat these people?
Mathematics
1 answer:
WITCHER [35]3 years ago
8 0

Answer:

Smallest number of rows that is needed to seat these people = 16

Step-by-step explanation:

Given - A theatre has seats arranged in rows of 24, three hundred eighty -two(382) people attend a show at the theatre.

To find - What is the smallest number of rows that is needed to seat these people?

Solution -

Given that,

Total number of seats in a Row = 24

Let us assume that, there are x rows

As.

There are 382 people in the Theatre

So,

Minimum seats required is

24 x = 382

⇒x = 382 ÷ 24

⇒x = 15.91

i.e.

Smallest number of rows that is needed to seat these people = 16

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Answer:

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Step-by-step explanation:

The equation of a line in point- slope form is

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where m is the slope and (a, b ) a point on the line

y + 3 = - \frac{4}{3} (x - 4) ← is in point- slope form

with slope m = - \frac{4}{3} and point on line (a, b ) = (4, - 3 )

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2 years ago
Erika worked 14 hours last week and 20 hours this week. If she earns $9 per hour, how much did she earn during these two weeks
IrinaK [193]
$9 per hour for 14 hours the first week can be found by doing 9 • 14, which equals 126.
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In the second week, Erika earned $180.
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3 years ago
Solve for x. Enter the solutions from least to greatest. (x−1)^2−9=0
Leno4ka [110]

Step-by-step explanation:

(x−1)²−9=0

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Read 2 more answers
The average expenditure on Valentine's Day was expected to be$100.89 (USA Today, February 13, 2006). Do male and femaleconsumers
stealth61 [152]

Answer:

(a) $62.16

(b) Male: $15.00

Female: $10.06

(c) Confidence Interval for male expenditure is ($106.40, $136.40)

Confidence interval for female expenditure is ($49.18, $69.30)

Step-by-step explanation:

(a) Male expenditure

Sample mean = $135.67, sd=$35, n=40, Z=2.576

Population mean = sample mean - (Z×sd)/√n = 135.67 - (2.576×35)/√40 = 135.67 - 14.27 = $121.40

Female expenditure

Sample mean= $68.64, sd=$20, n=30, Z=2.576

Population mean = 68.64 - (2.576×20)/√30 = 68.64 - 9.40 = $59.24

$121.40 - $59.24 = $62.16

(b) Male: Error margin = (t-value × sd)/√n

Degree of freedom = n-1 = 40-1= 39. t-value corresponding to 39 degrees of freedom and 99% confidence level is 2.708

Error margin = (2.708×35)/√40 = 94.78/6.32 = $15.00

Female

Degrees of freedom = n-1 = 30-1 = 29. t-value is 2.756

Error margin = (2.756×20)/√30 = 55.12/5.48 = $10.06

(c) Male

Confidence Interval (CI) = (mean + or - error margin)

CI = 121.4 + 15.00 = $136.40

CI = 121.4 - 15.00 = $106.40

Confidence Interval is ($106.40, $136.40)

Female

CI = 59.24 + 10.06 = $69.30

CI = 59.24 - 10.06 = $49.18

Confidence Interval is ($49.18, $69.30)

7 0
4 years ago
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