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Lady bird [3.3K]
2 years ago
11

Jackie won tickets playing the bowling game at the local arcade. The first time, she won 60 tickets. The second time, she won a

bonus, which was 4 times the number of tickets of the second prize. Altogether she won 200 tickets. How many tickets was the second prize?
Mathematics
1 answer:
Fynjy0 [20]2 years ago
4 0

Answer: i'm pretty sure she got 35 the second time. hope this helps

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Check all of the ordered pairs that satisfy the equation below.y=2/5x
zvonat [6]

We are going to do this step by step:

Let's start with option A

A.

X = 25 and Y = 5/2

y=\frac{2}{5}\cdot x=\frac{2}{5}\cdot25=\frac{2\cdot25}{5}=\frac{50}{5}=\frac{5\cdot10}{5\cdot1}=10

In this case, when X = 25 , then Y = 10 ,which is different to 5/2

B.

X = 14 , Y = 35

y=\frac{2}{5}\cdot14=\frac{28}{5}

In case, when X = 14, then Y = 28/5, which is different to 35

C.

X = 40 , Y = 24

y=\frac{2}{5}\cdot40=\frac{80}{5}=16

Similarly, in this case, when X = 40, then Y = 16, which is different to 24

D.

X = 10 , Y=4

y=\frac{2}{5}\cdot10=\frac{20}{5}=4

Now, in this case, we can see that when X = 10, then Y = 4 which is the same as the given value of Y

E.

X = 50 , Y = 20

y=\frac{2}{5}\cdot50=\frac{100}{5}=20

In this case, the values of Y are also the same.

F.

X = 30 , Y = 12

y=\frac{2}{5}\cdot30=\frac{60}{5}=12

Again, in this case, the values of Y are the same, so the pair satisfies the equation.

In conclusion: options D, E and F satisfy the equation

3 0
11 months ago
Can you guys help me on this
WITCHER [35]

Answer: 1/4 OR 1/8

Step-by-step explanation:

the first coin can be heads which has a 1/2 chance of happening and the other coin also has a chance to get tails which is 1/2

if you add the 1/2 and the other 1/2 you can get 1/4 OR 1/8

4 0
2 years ago
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NO LINKS!!<br><br>Solve each equation. Show all work.<br><br>(x - 2)^2 - 64 = 0​
klasskru [66]

(x-2)^2-64=0\\(x-2)^2=64\\x-2=8 \vee x-2=-8\\x=10 \vee x=-6

4 0
2 years ago
Read 2 more answers
Question down here below ?
Sergio [31]

Answer:

1 i doing well crewmates have you seen the impasta yet?

2 you have known well

3 it will be like that sometimes.

Step-by-step explanation:

It is what it is.

3 0
2 years ago
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Two 6-sided dice are tossed. One die is red and the other is white, so that they are distinguishable. (That is, we consider the
Darina [25.2K]

Answer:

Given the following events and its elements when two 6-sided dice are tossed:

A: the sum of the dice is even

B: at least one die shows a 3

C: the sum of the dice is 7

The elements of the intersections are:

a) A∩B={(1, 3),(3, 1),(3, 3),(3, 5),(5, 3)}

b) B^c∩C={(1, 6),(2, 5),(5, 2),(6, 1)}

c) A∩C={∅}

d) A^c∩B^c∩C^c={(1, 2),(1, 4),(2, 1),(4, 1),(4, 5),(5, 4),(5, 6),(6, 5)}

Step-by-step explanation:

The total number of elements of the universal set (U) for this problem is 36 elements because the number of possible combinations is 6*6.  

For the event A, half of the elements satisfy the condition of the sum being an even number.

A={(1, 1),(1, 3),(1, 5),...,(6, 2),(6, 4),(6, 6)}=18 elements

For event B, the elements that contain a 3 are:

B={(1, 3),(2, 3),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 3),(5, 3),(6, 3)}= 11 elements

For event C, the sum of the elements is 7:

C={(1, 6),(2, 5),(3, 4),(4, 3),(5, 2),(6, 1)}=6 elements

Now let's find the intersections:

a) A∩B are the elements of A that have a 3.

A∩B={(1, 3),(3, 1),(3, 3),(3, 5),(5, 3)}

b) B^c∩C are the elements of the universal set (U) that do not have a 3 and that the sum of the dice is 7

B^c∩C={(1, 6),(2, 5),(5, 2),(6, 1)}

c) A∩C are the elements of that sum 7, but this is not possible given that all the elements of A sum an even number and 7 is not an even number.

A∩C={∅}

d) A^c∩B^c∩C^c are the elements that don't sum an even number, don't have a 3 and the sum is not 7.

A^c∩B^c∩C^c={(1, 2),(1, 4),(2, 1),(4, 1),(4, 5),(5, 4),(5, 6),(6, 5)}

5 0
2 years ago
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