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Mandarinka [93]
1 year ago
8

Mario and Bowser begin with a hat that has 5 balls in it: 3 are red, 2 are green. Bowser secretly picks a ball and tells Mario i

t is red, although he might be lying about the color. If there is an 80% chance Bowser is lying, what is the probability that Mario draws 2 red balls in a row after Bowser's initial draw?
Mathematics
1 answer:
dangina [55]1 year ago
5 0
20% probability. because 100-80=20 so therefore it would be 20%
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four times the first of three consecutive odd integers is five less than the product of three and the third integer find the int
Andre45 [30]
I don't understand what integer is being asked for. The question is poorly worded.

The three integers are 7, 9, 11.
4 times 7 = 28
and 3 times 11 = 33
and 28 is 5 less than 33.
8 0
1 year ago
Jung-Soon has $25 to spend on prizes for a game at the school fair. Lip balm costs $1.25 each, and
Dvinal [7]

Answer:

1.25x + 1.5y=25

Step-by-step explanation:

let the number of lip balms that can be bought be x and the notebooks be y then their total cost is 1.25x + 1.5y. The total $25 there for the equation becomes 1.25x + 1.5y=25.

6 0
1 year ago
The final velocity, V, of an object under constant acceleration can be found using the formula uppercase V squared = v squared +
Anon25 [30]

Answer:

StartFraction uppercase V squared minus v squared Over 2 s EndFraction = a

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
A ball is thrown upwards and it’s height h m,at time t s,is given by h=7t-4.9t2.How long does the ball spend more than 1.5m abov
Allushta [10]

Answer:  From what you wrote the I think the ball is drop

At time 0 you are 7 meters above ground

You basically just need to solve for

1.5 =7t-4.9t^2

by rearrange the equation you get

4.9t^2-7t+1.5=0

by using Quadratic Formula

t=1.166s

4 0
1 year ago
Two armadillos and three aardvarks sat in a five-seat row at the Animal Auditorium. What is the probability that the two animals
Serhud [2]

Answer:

      \dfrac{1}{15}

Explanation:

The number of different ways in which the<em> two armadillos</em> would be at the ends of the row is 2:

                      P(2,2)=\dfrac{2!}{(2-2)!}=\dfrac{2}{0!}=2

The number of different combinations in which<em> two of the three aardvarks </em>can sit at the ends of the row is P(3,2):

          P(3,2)=\dfrac{3!}{(3-2)!}=\dfrac{3\times 2\times 1}{1}=6

Therefore, there are 2 + 6 = 8 different ways in which the two animlas on the ends of the row were both armadillos or both aardvarks.

Now calculate the total number of different ways in which the animals can sit. It is P(5,5):

        P(5,5)=\dfrac{5!}{(5-5)!}=5!=5\times 4\times 3\times 2\times 1=120

Thus, <em>the probability that the two animals on the ends of the row were both armadillos or both aardvarks</em> is equal to the number of favorable outputs divided by the total number of possible outputs:

      Probability=\dfrac{8}{120}=\dfrac{1}{15}

8 0
1 year ago
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