I need the answer really soon and accurate.

a conference hall has 33 rows.the last row contains 80 seats each row has two fewer seats than the row behind it.how many seats

ella [17]

Answer:

there are 84 seats in the second to the last row

Step-by-step explanation:

Given that

each row has two fewer seats than the row behind it

It means that

if first row has x seats then

no of seat in second row has (x-2) seats.

and similarly

no of seat in third row =no of seat in second row - 2seats.

no of seat in third row = (x-2) seat - 2 seats = x - 4 seats

_____________________________________________

Given that

last row has 80 seats

let the number of seat in first to the last row be x

then number of seat in last row = x-2

we know number of seat in last row = 80

thus

x - 2 = 80

x = 82

Thus,

the number of seat in first to the last row is 82

Similarly let the number of seat in second to the last row be y

then number of seat in first to the last row = y-2

since number of seat in first to the last row is 82

thus

y-2 = 82

y = 82 + 2 = 84

Thus, there are 84 seats in the second to the last row.

_________________________________________________

8 0

4 years ago

How do I graph this solution using linear inequalities

JulijaS [17]

Im not sure if this helpful or not but I hope that it is ^-^
im sorry if its not

6 0

3 years ago

Which model shows 25%?

Stels [109]

Answer:

The first one shows 25%

In the 1st modal, a whole is divided into 4 and 1 of the 4 is selected or highlighted

And as the whole rectangle is 100% and being divided into four its becomes 100/4% and when we solve we get 100/4=25

And thus its the first modal that shows 25%

In the second case its divide into 3 boxes so.. 100/3% and when we solve we get 33.3333333.......and this is not what we want

THEREFORE, THE CORRECT ANSWER IS 1ST MODAL

8 0

3 years ago

A beverage is made by mixing 2 parts of water with 3 parts of fruit juice. How many parts of water are mixed with 1 part of frui

photoshop1234 [79]

2/3 = x/1...2 parts water to 3 parts juice = x parts water to 1 part juice
cross multiply
(3)(x) = (2)(1)
3x = 2
x = 2/3 parts water are needed for 1 part juice

6 0

4 years ago

Suppose we have 14 red balls and 14 green balls as in the previous exercise. Show that at least two pairs, consisting of one red

Nuetrik [128]

Answer:

since each ball has a different number and if no two pairs have the same value there is going to be 14∗14 different sums. Looking at the numbers 1 through 100 the highest sum is 199 and lowest is 3, giving 197 possible sums

For the 14 case, we show that there exist at least one number from set {3,4,5,...,17} is not obtainable and at least one number from set {199,198,...,185} is not obtainable.

So we are left with 197 - 195 options

14 x 14 = 196

196 > 195

so there are two pairs consisting of one red and one green ball that have the same value

As to the comment, I constructed a counter-example list for the 13 case as follows. The idea of constructing this list is similar to the proof for the 14 case.

Red: (1,9,16,23,30,37,44,51,58,65,72,79,86)

Green: (2,3,4,5,6,7,8;94,95,96,97,98,99,100)

Note that 86+8=94 and 1+94=95 so there are no duplicated sum

Step-by-step explanation:

For the 14 case, we show that there exist at least one number from set {3,4,5,...,17} is not obtainable and at least one number from set {199,198,...,185} is not obtainable.

First consider the set {3,4,5,...,17}.

Suppose all numbers in this set are obtainable.

Then since 3 is obtainable, 1 and 2 are of different color. Then since 4 is obtainable, 1 and 3 are of different color. Now suppose 1 is of one color and 2,3,...,n−1 where n−1<17 are of the same color that is different from 1's color, then if n<17 in order for n+1 to be obtainable n and 1 must be of different color so 2,3,...,n are of same color. Hence by induction for all n<17, 2,3,...,n must be of same color. However this means there are 16−2+1=15 balls of the color contradiction.

Hence there exist at least one number in the set not obtainable.

We can use a similar argument to show if all elements in {199,198,...,185} are obtainable then 99,98,...,85 must all be of the same color which means there are 15 balls of the color contradiction so there are at least one number not obtainable as well.

Now we have only 195 choices left and 196>195 so identical sum must appear

A similar argument can be held for the case of 13 red balls and 14 green balls

6 0

4 years ago