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Leokris [45]
3 years ago
15

A producer has fixed expenses of $980 to put on a 3-day production of a play. Each ticket to the play will be sold for $5.00. Th

e line shows the net profit or loss of the producer based on selling x tickets. What is the minimum number of tickets the producer must sell over the 3-day period in order to earn a profit?
Mathematics
1 answer:
zysi [14]3 years ago
3 0

Answer:

196 tickets

Step-by-step explanation:

The total expenses to put on a 3-day production of a play is $980. Each ticket to go and watch the play is going to be sold for $5. The producer sells x number of tickets, x number of tickets would cost 5x. Therefore to find the number of tickets the producer must sell so as to make a profit the money generated from selling the tickets must be more than the cost of production. Therefore we use the formula:

5x ≥ 980

Dividing through by 5:

x ≥ 980

x ≥ 980/5

x ≥ 196

Therefore the producer must sell at least 196 tickets so as to make a profit

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babymother [125]

Answer:

What we know is that 30% of the group have brown hair and are girls. It means that in the other 70% we have the boys with brown hair, the other boys and the other girls.

In the case all the girls have brown hair, the 70% remaining would be boys. So 70% of then would have brown hair. (0.7*0.7=0.49 -> 49% of the children would be brown hair) so 49% (boys) and 30% (girls) are more than 50%..

Now, let's suppouse that 90% of the group are girls. It means 10% are boys and 70% of that 10% (=7%) are the boys in the group that have brown hair. Now 7% + 30% of the children have brown hair, wich isn't more than the half.

We have explained with two extreme cases that the information is not enough for making a precise answer

7 0
2 years ago
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Anuta_ua [19.1K]

Answer:

\sin(105) = \frac{\sqrt 2 + \sqrt 6}{4}

Step-by-step explanation:

Given

\sin(105^o)

Required

Solve

Using sine rule, we have:

\sin(A + B) = \sin(A)\cos(B) + \sin(B)\cos(A)

This gives:

\sin(105^o) = \sin(60 + 45)

So, we have:

\sin(60 + 45) = \sin(60)\cos(45) + \sin(45)\cos(60)

In radical forms, we have:

\sin(60 + 45) = \frac{\sqrt 3}{2} * \frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} * \frac{1}{2}

\sin(60 + 45) = \frac{\sqrt 6}{4} + \frac{\sqrt 2}{4}

Take LCM

\sin(60 + 45) = \frac{\sqrt 6 + \sqrt 2}{4}

Rewrite as:

\sin(60 + 45) = \frac{\sqrt 2 + \sqrt 6}{4}

Hence:

\sin(105) = \frac{\sqrt 2 + \sqrt 6}{4}

3 0
2 years ago
I need help please Solve the problems! CHECK PICTURE FOR QUESTION
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Check the picture below.

we know that AL is an angle bisector, so the angle at A gets cuts into two equal halves, we also know the angle at B is 30°, so the triangle ABC is really a 30-60-90 triangle, meaning the angle at A is really a 60° angle, cut in two halves gives us 30° and 30° as you see in the picture.

if the angles at A and B, inside the triangle ABL, are equal, twin angles are only made in an isosceles by twin sides, that means that AL = BL.

Looking at the triangle ALC, we can see is also another 30-60-90 triangle, and we can just use the 30-60-90 rule to get x=CL.

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