Answer:
1806 seats.
Step-by-step explanation:
From the question given above, the following data were obtained:
Row 1 = 24 seats
Row 2 = 27 seats
Row 3 = 30 seats
Total roll = 28
Total number of seat =?
From the above data, we can liken the roll to be in arithmetic progress.
Also, we are asked to determine the total number of seats in the theater.
Thus the sum of the sequence can be written as:
Roll 1 + Roll 2 + Roll 3 +... + Roll 28 i.e
24 + 27 + 30 +...
Thus, we can obtain obtained the total number of seats in the theater by applying the sum of arithmetic progress formula. This can be obtained as follow:
First term (a) = 24
Common difference (d) = 2nd term – 1st term
Common difference (d) = 27 – 24 = 3
Number of term (n) = 28
Sum of the 28th term (S₂₈) =?
Sₙ = n/2 [2a + (n –1)d]
S₂₈ = 28/2 [2×24 + (28 –1)3]
S₂₈ = 14 [48 + 27×3]
S₂₈ = 14 [48 + 81]
S₂₈ = 14 [129]
S₂₈ = 1806
Thus, the number of seats in the theater is 1806.
Answer:
(x + 9) (x - 8)
Replace a with 9
and b with -8
Step-by-step explanation:
Simple, just divide the number by the percentage and you get the original number. You can check it by multiplying the original number of 272 by 55% or .55 to get 150.
150/.55 = 272
Answer:
$200
Step-by-step explanation:
Here is the full question
You can upgrade lighting at your factory to LED bulbs that cost $6.95 each and last an average of 5 years. It costs $3 in labor to change a bulb. Over a 10-year period, about how much will it cost per year to install LED bulbs in 100 lamps and change the bulbs when they burn out?
in 10 years, the bulbs would be changed twice
Total cost incurred per bulb in 10 years = (6.95 + $3) x 2 = $19.90
Since there are 100 bulbs, total cost incurred would be - $19.90 x 100 = $1990
Cost per year = $1990 / (6.95 + $3) = $200