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TiliK225 [7]
3 years ago
10

What is 2+3+4-5%= What is the anwser?

Mathematics
1 answer:
Gnoma [55]3 years ago
8 0

Answer:179 /20

(Decimal: 8.95)

Step-by-step explanation:

2+3+4− 5 /100

=5+4− 5 /100

=9− 5 /100

=9− 1 /20

= 179 /20

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Arrange the hyperbolas in increasing order of the horizontal widths of their asymptote rectangles.
almond37 [142]

The  increasing order of the horizontal widths of their asymptote rectangles is dependent on the values gotten from  y = ± x.

<h3>What is a Hyperbola?</h3>

This is defined as a two-branched open curve formed by the intersection of a plane perpendicular to the bases of a double cone.

The rectangular hyperbola has two asymptotes which are defined as y = ± x in this scenario.

Read more about Hyperbola here brainly.com/question/3351710

#SPJ1

8 0
2 years ago
2d(3*6)=48y when y=1/4
vaieri [72.5K]

Answer: d = 1/3

Step-by-step explanation: 48y = 48/4 = 12. So, 2d(18) is equal to 12. 12/18 is 2/3 so 2d has to equal 2/3. That means d = 1/3.

8 0
2 years ago
(3x + 1/2 ) + ( 7x - 4 1/2 ) simply the expression
ololo11 [35]

(3x + 1/2 ) + ( 7x - 4 1/2 )

3x+7x + 1/2 - 4 1/2

10x -4

5 0
4 years ago
1. A football team charges $30 per ticket and averages 20,000 people per game. Each
IRINA_888 [86]

Question: A Football Team Charges $30 Per Ticket And Averages 20,000 People Per Game. Each Person Spends An Average Of $8 On Concessions. For Every Drop Of $1 In Price, The Attendance Rises By 800 People. What Ticket Price Should The Team Charge To Maximize Total Revenue? Calculate The TR Max.

This problem has been solved!

See the answer

A football team charges $30 per ticket and averages 20,000 people per game. Each person spends an average of $8 on concessions. For every drop of $1 in price, the attendance rises by 800 people. What ticket price should the team charge to maximize total revenue? Calculate the TR max.

$50

                                                                                                                                             

7 0
3 years ago
Six​ stand-up comics,​ A, B,​ C, D,​ E, and​ F, are to perform on a single evening at a comedy club. The order of performance is
kiruha [24]

Answer:

a) 1/6

b) 1/36

c) 1/720

d) 1/3

Step-by-step explanation:

a) Any of the six comics can perform in the fourth place, so there is one chance in six that Comic F is the one that performs fourth.

P(F=4th)=1/6

b) In this case, we have two conditions. Both are independent of each other, so the probability of both happening is the product of the probabilities of each happening individually:

P(D=4th \&B=2nd)=P(D=4th)\cdot P(B=2nd)=(1/6)\cdot(1/6)=1/36

c) This combination is one in all possible orders of perform. The amount of combinations of orders is n!=6!=720 possible combinations. So the probability of this specific order is:

P(F, D, B, A, C, E)=1/720

d) In this case, of the six possible comics performing in the last place, we calculate the probability of 2 of them being in that place. So the probability is:

P(C=6th\,or\, E=6th)=2/6=1/3

8 0
3 years ago
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