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VMariaS [17]
3 years ago
10

HELP ASAP!!! HELP FAST PLEEASE!!!

Mathematics
1 answer:
expeople1 [14]3 years ago
8 0

Answer:

e.

Step-by-step explanation:

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You need 3 tickets for one go-cart ride .How many tickets do you need for 5 go-cart rides
Sphinxa [80]
3 tickets for 1
? tickets for 5
1 has been multiplied by 5 here so do the same to the 3 and you get 15.
15 tickets for 5 rides
4 0
3 years ago
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it rained nine days in the month of april .based on this ,what is the probaility that is does not rain on the first day of may?​
attashe74 [19]
<h3>Answer:     7/10</h3>

==========================================================

Explanation:

There are 30 days in April. Since it rained 9 of those days, the empirical probability of it raining in April is 9/30 = (3*3)/(3*10) = 3/10.

If we assume that the same conditions (ie weather patterns) hold for May, then the empirical probability of it raining in May is also 3/10. By "raining in May", I mean specifically raining on a certain day of that month.

The empirical probability of it not raining on the first of May is therefore...

1 - (probability it rains)

1 - (3/10)

(10/10) - (3/10)

(10-3)/10

7/10

We can think of it like if we had a 10 day period, and 3 of those days it rains while the remaining 7 it does not rain.

6 0
3 years ago
A carnival charges two different prices for admissions. Adults cost $4 while children cost $1.50. If a total of $5050 was collec
GenaCL600 [577]

Answer:

Step-by-step explanation:

We have to have 2 different equations to solve this.  One equation will represent the number of tickets sold while the other represents the money collected when the tickets were sold.

We know that adult tickets + children tickets = 2200 tickets.

That's the "number of tickets" equation.  Let's call adult tickets "a" and children's tickets "c".  So a + c = 2200

Now if each adult costs $4, then the expression that represents that as a cost is 4a.  If there is 1 adult, the cost is $4(1) = $4; if there are 2 adults, the cost is $4(2) = $8; if there are 3 adults, the cost is $4(3) = $12, etc.

The same goes for the children's tickets.  If each child's ticket is $1.50, then the expression that represents the cost of a child's ticket is 1.5c (we don't need the 0 at the end; it doesn't change anything to drop it off).  The total money brought in from the cost of these tickets was $5050, so

4a + 1.5c = 5050

Let's solve the first equation for a.  If a + c = 2200, then a = 2200 - c.  Sub that into the second equation and solve it for c:

4(2200 - c) + 1.5c = 5050 and

8800 - 4c + 1.5c = 5050 and

-2.5c = -3750 so

c = 1500

That means that there were 1500 children's tickets sold.  If a + c = 2200, then a + 1500 = 2200 so

a = 2200 - 1500 so

a = 700

There were 1500 children's tickets sold and 700 adult tickets sold.

7 0
3 years ago
NEED HELP ASAPPP
Serjik [45]
The answers is a because the radius must be 5
6 0
3 years ago
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Renee is wanting to join a gym.
WINSTONCH [101]
Pumping iron: 10x+90=0
Fit for life: 15x+50=0

Set them equal to each other. 10x+90=15x+50. Solve for x. Subtract 50 from both sides, get 10x+40=15x, then subtract 10x from both sides, getting 40=5x. Divide 5 from both sides to isolate x, x=8. After 8 months the two gyms will cost the same.
7 0
3 years ago
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