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viktelen [127]
3 years ago
11

At a concert, 20% of the audiences is sitting in the balcony. If there are 450 people sitting in the balcony, what is the total

number of people at the concert?
Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
3 0
....................
aksik [14]3 years ago
3 0
90 peeps at the balcony
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You and five friends have joined baseball.The team membership fee is $50 plus a $5 fee per game to pay the referee.
PtichkaEL [24]
275 beacues if u and 50 five times it would of been 250 but u an the five two the 50 and u and 55 five times and u get 275
7 0
3 years ago
Y = 2/3 when x = 0.2
azamat
Please explain some more:)
4 0
3 years ago
Find the area of the rhombus.<br> 4.32 m2<br> 2.16 m2<br> 1.08 m2<br> 1.05 m2
S_A_V [24]

Answer:

Step-by-step explanation:

Alright, lets get started.

The part of the diagonals are given as 0.9 m and 1.2 m.

The diagonals of rhombus intersect each other.

Hence the length of diagonals will be : 0.9*2=1.8 and 1.2*2=2.4

The formula of area of rhombus is : \frac{pq}{2} where p and q are the length of diagonals.

So plugging the values of diagonals in formula, the area will be :

Area = \frac{1.8*2.4}{2}

Area = \frac{4.32}{2}

Area = 2.16 \ m^2   ................. Answer

Hope it will help :)

6 0
3 years ago
A coin is tossed 600times with the frequencies as:
brilliants [131]

Answer:

  • see below

Step-by-step explanation:

Total number of trials =600.

Number of heads = 342.

Number of tails = 258

On tossing a coin,

let  \: E_1  \: and  \: E_2  \: be  \: the  \: events \:  of \:  getting  \: a  \: head

and of getting a tail respectively.

then,

1)  \: P \:  (getting \:  a  \: head) = \: P \:  (E_1)

\longrightarrow \:  \frac{number \: of \: heads \: coming \: up}{total \: number \: of \: trials}

\longrightarrow \:  \frac{342}{600}  =  \frac{57}{100}

\longrightarrow \: 0.57

\red{ \rule{150pt}{3pt}} \:

2) P (getting \: a\: tail) =P (E_2)

\longrightarrow \:  \frac{number \: of \: tails \: coming \: up}{total \: number \: of \: trials}

\longrightarrow \:  \frac{258}{600}  =  \frac{43}{100}

\longrightarrow \: 0.43

6 0
3 years ago
A dairy company gets milk from two dairies and then blends the milk to get the desired amount of butterfat. Milk from dairy I co
zubka84 [21]

Answer:

a) i The company should buy 40 gallons from dairy I and 60 gallons from dairy

ii) What is the maximum amount of​ butterfat? The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b.The excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

Step-by-step explanation:

a. How much milk from each supplier should the company buy to get at most 100 gallons of milk with the maximum amount of​ butterfat?

From the question, we are told that:

Milk from dairy I costs ​$2.40 per​ gallon, Milk from dairy II costs ​$0.80 per gallon.

Let's represent:

Number of gallons of Milk from dairy I = x

Number of gallons of Milk from dairy II = y

At most ​$144 is available for purchasing milk.

$2.40 × x + $0.80 × y = 144

2.40x + 0.80y = 144........ Equation 1

x + y = 100....... Equation 2

x = 100 - y

2.40(100 - y) + 0.80y = 144

240 - 2.4y + 0.80y = 144

-1.60y = 144 - 240

-1.6y = -96

y = -96/-1.6

y = 60

From Equation 2

x + y = 100....... Equation 2

x + 60 = 100

x = 100 - 60

x = 40

Therefore, since number of gallons of Milk from dairy I = x and number of gallons of Milk from dairy II = y

The company should buy 40 gallons from dairy I and 60 gallons from dairy

II. What is the maximum amount of​ butterfat?

From the question

Dairy I can supply at most 50 gallons averaging 3.9​% ​butterfat,

50 gallons = 3.9% butterfat

40 gallons =

Cross Multiply

= 40 × 3.9/50

= 3.12%

Dairy II can supply at most 90 gallons averaging 2.9​% butterfat.

90 gallons of milk = 2.9% butter fat

60 gallons of milk =

Cross Multiply

= 60 × 2.9%/90

=1.9333333333%

≈ 1.93%

The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b. The solution from part a leaves both dairy I and dairy II with excess capacity. Calculate the amount of additional milk each dairy could produce.

Excess capacity of Diary I =

50 gallons - 40 gallons = 10 gallons

Excess capacity of Diary II =

90 gallons - 60 gallons = 30 gallons

Therefore, the excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

3 0
3 years ago
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