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galina1969 [7]
1 year ago
12

In a recent year, team 1 made 191 out of 238 free- throw attempts and team 2 made 106 out of 160 free- throw attempts. Copy and

use the percent bars to answer each question.

Mathematics
2 answers:
ivann1987 [24]1 year ago
5 0
Cross multiply and divide Put 191 over 238 (like a fraction) the put another fraction bar right next to your first fraction and put 100 on bottom (this equals a whole in % and it's right across from the total of 238 free throws) then multiply 191 by 100 then divide by 238 and that answer should be your the percent you need.
IRISSAK [1]1 year ago
4 0

Answer:

Team A80.25%

Team B66.25%

find the graphical representation as attached

Step-by-step explanation:

In a recent year, team 1 made 191 out of 238 free- throw attempts and team 2 made 106 out of 160 free- throw attempts. Copy and use the percent bars to answer each question.

lest find the percentage throw of team A

191/238*100%

80.25%

the percentage throw of team B is also

106/160*100%

66.25%

find the graphical representation as attached below.

the vertical side represents the percentage throw of team 1 and 2, while the horizontal side is the teams

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Which system of equations can be used to find the roots of the equation 4 x Superscript 5 Baseline minus 12 x Superscript 4 Base
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3 0
1 year ago
On a certain​ route, an airline carries 7000 passengers per​ month, each paying ​$30. A market survey indicates that for each​ $
KengaRu [80]

Answer:

The ticket price that maximizes revenue is $50.

The maximum monthly revenue is $250,000.

Step-by-step explanation:

We have to write a function that describes the revenue of the airline.

We know one point of this function: when the price is $30, the amount of passengers is 7000.

We also know that for an increase of $1 in the ticket price, the amount of passengers will decrease by 100.

Then, we can write the revenue as the multiplication of price and passengers:

R=p\cdot N=(30+x)(7000-x)

where x is the variation in the price of the ticket.

Then, if we derive R in function of x, and equal to 0, we will have the value of x that maximizes the revenue.

R(x)=(30+x)(7000-100x)=30\cdot7000-30\cdot100x+7000x-100x^2\\\\R(x)=-100x^2+(7000-3000)x+210000\\\\R(x)=-100x^2+4000x+210000\\\\\\\dfrac{dR}{dx}=100(-2x)+4000=0\\\\\\200x=4000\\\\x=4000/200=20

We know that the increment in price (from the $30 level) that maximizes the revenue is $20, so the price should be:

p=30+x=30+20=50

The maximum monthly revenue is:

R(x)=(30+x)(7000-100x)\\\\R(20)=(30+20)(7000-100\cdot20)\\\\R(20)=50\cdot5000\\\\R(20)=250000

3 0
2 years ago
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