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Leto [7]
3 years ago
13

At the city museum, child admission is $5.30 and adult admission is $9.40 . On Tuesday, 140 tickets were sold for a total sales

of $1086.40 . How many child tickets were sold that day?
Mathematics
2 answers:
7nadin3 [17]3 years ago
7 0
Total tickets sold = # of kid's tickets sold + # of adult tickets sold
Total money made = amount of money kid's tickets raised + amount the adult tickets made

These can be rephrased as
140 = k + a
and
$1086.40 = $5.30k + $9.40a

Rephrase the first;
140 = k + a
a = 140 - k

Plug the equation into the first;
1086.4 = 5.3k + 9.4a
1086.4 = 5.3k + 9.4(140 - k)
1086.4 = 5.3k + 1316 - 9.4k
-229.6 = -4.1k
<span>k = 56 tickets</span>
sergeinik [125]3 years ago
5 0
C + a = 140......a = 140 - c
5.3c + 9.4a = 1086.4

5.3c + 9.4(140 - c) = 1086.4
5.3c + 1316 - 9.4c = 1086.4
5.3c - 9.4c = 1086.4 - 1316
-4.1c = - 229.6
c = -229.6 / -4.1
c = 56......so their were 56 child tickets sold
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Given, that the librarian charges $0.30 per each day that the book is late to submit

for estimating the relation between the number of days the book is late and the total fine fee

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so from above, it is clear that the linear equation will be

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The equivalent equation to find the ordered pair will be y=0.3x

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4 0
1 year ago
I need help on this what is RS
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11

Step-by-step explanation:

If RS=TV

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3x=12

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2 years ago
Each ticket to enter the zoo costs 7$ adam bought two bags of peanuts for 4$ and one of his brothers bought a lion poster for 12
Vika [28.1K]
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5 0
3 years ago
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At a certain pizza parlor, % of the customers order a pizza containing onions, % of the customer's order a pizza containing saus
wel

Answer:

At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing sausage, and 66% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.

Step-by-step explanation:

Hello!

You have the following possible pizza orders:

Onion ⇒ P(on)= 0.36

Sausage ⇒ P(sa)= 0.35

Onions and Sausages ⇒ P(on∪sa)= 0.66

The events "onion" and "sausage" are not mutually exclusive, since you can order a pizza with both toppings.

If two events are not mutually exclusive, you know that:

P(A∪B)= P(A)+P(B)-P(A∩B)

Using the given information you can use that property to calculate the probability of a customer ordering a pizza with onions and sausage:

P(on∪sa)= P(on)+P(sa)-P(on∩sa)

P(on∪sa)+P(on∩sa)= P(on)+P(sa)

P(on∩sa)= P(on)+P(sa)-P(on∪sa)

P(on∩sa)= 0.36+0.35-0.66= 0.05

I hope it helps!

5 0
3 years ago
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katen-ka-za [31]

Answer:

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x^2=338

\sqrt{x^2} =\sqrt{338}

x=\sqrt{338}

8 0
3 years ago
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