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3 years ago

10

Tickets to a football final are selling well. On Thursday, 47 of the tickets are sold. On Friday, 14 of the tickets are sold. Wh

at fraction of tickets are available to sell on Saturday?

1 answer:

lbvjy [14]3 years ago

6 0

The question seems incomplete ; as the total number of tickets to be sold isn't given.

Answer:

61 / X

Step-by-step explanation:

Let's take the total Number of tickets to be sold as : X

Number of tickets sold on Thursday = 47

Number sold on Friday = 14

Fraction of tickets available for sale on Saturday :

(Total number of tickets already sold) / Total number of tickets to be sold

(Thursday + Friday sales) / total number of tickets to be sold

Fraction available for sale on Saturday = (47+14) / X

Fraction available for sale on Saturday = 61 / X

Kindly put value of x = total number of tickets available for sale to get the exact fraction.

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Pls solve the simultaneous equation in the attachment.

siniylev [52]

Answer:

Part a) The solution is the ordered pair (6,10)

Part b) The solutions are the ordered pairs (7,3) and (15,1.4)

Step-by-step explanation:

Part a) we have

$\frac{x}{2}-\frac{y}{5}=1$ ----> equation A

$y-\frac{x}{3}=8$ ----> equation B

Multiply equation A by 10 both sides to remove the fractions

$5x-2y=10$ ----> equation C

isolate the variable y in equation B

$y=\frac{x}{3}+8$ ----> equation D

we have the system of equations

$5x-2y=10$ ----> equation C

$y=\frac{x}{3}+8$ ----> equation D

Solve the system by substitution

substitute equation D in equation C

$5x-2(\frac{x}{3}+8)=10$

solve for x

$5x-\frac{2x}{3}-16=10$

Multiply by 3 both sides

$15x-2x-48=30$

$15x-2x=48+30$

Combine like terms

$13x=78$

$x=6$

<em>Find the value of y</em>

$y=\frac{x}{3}+8$

$y=\frac{6}{3}+8$

$y=10$

The solution is the ordered pair (6,10)

Part b) we have

$xy=21$ ---> equation A

$x+5y=22$ ----> equation B

isolate the variable x in the equation B

$x=22-5y$ ----> equation C

substitute equation C in equation A

$(22-5y)y=21$

solve for y

$22y-5y^2=21$

$5y^2-22y+21=0$

Solve the quadratic equation by graphing

The solutions are y=1.4, y=3

see the attached figure

<em>Find the values of x</em>

For y=1.4

$x=22-5(1.4)=15$

For y=3

$x=22-5(3)=7$

therefore

The solutions are the ordered pairs (7,3) and (15,1.4)

3 0

4 years ago

Here are five number cards. 2 5 7 8 9 One of the cards is removed and the mean average of the remaining four number cards is 6.

Lubov Fominskaja [6]

Answer:

Card 7

Step-by-step explanation:

2+5+7+8+9 =31

Multiples of 6: 6, 12, 18, 24, 30

31-9=22, No

31-8=23, No

31-5=26, No

31-2=29, No

31-7=24, Yes

The only way to get a multiple of 6 is to subtract card 7, so card 7 is the answer.

7 0

3 years ago

A cake has circumference of 25 1/7 inches. What is the area of the cake? Use 22/7 to approximate π. Round to the nearest hundred

choli [55]

Answer:

50.29 in^2

Step-by-step explanation:

From the circumference we need to find the radius.

Since C = 2*pi*r, r = C / (2*pi).

Here, with C = 25 1/7 in, r = (25 1/7 in) / [ 2(22/7) ], or r = 4 in

The area of the cake is A = pi*r^2, or (here) A = (22/7)(4 in)^2 = 50.29 in^2

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3 years ago

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love history [14]

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4 years ago

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ki77a [65]

I think that your answer is 32

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