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zaharov [31]
4 years ago
8

Ind the perimeter and area of the sandbox. Make sure your answers have the correct number of significant digits. A children's sa

ndbox measures 7.8 feet by 5.23 feet.
Mathematics
1 answer:
lawyer [7]4 years ago
8 0
Perimeter: 26.06ft

area: 40.794 
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scZoUnD [109]

Answer:

Although, my life can't end here, life has so much more to offer me. I collected the last of my strength to swim up to to surface where I can be released from the horrors of the deep waters. I swallowed gulps of the chlorine water and my eyes ached from being open for to long. I wrestled the clenching fists of the water that urged me to stay with them. I swam, and swam, swam, until I reached my hands out for help, and that's when I knew I made it. I could feel hands from above grasping me and helping me climb out of the bitter cold pool. I ran into my mother's arms, as a gush of relief spread through my body. The relief of staying in my mothers arm, that helped me forget the isolated pits of the dark waters.

Sorry if it's not that good! I'm not the best writer, and you don't have to use this ending paragraph if you don't want to!!

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Step-by-step explanation:

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Ethan says that the value of 40.7 is 10 times the value of 4.07 is be correct?
VLD [36.1K]

Answer:

He is correct

Step-by-step explanation:

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A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a
Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

                                   p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

p^4 (1-p)

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

(1-p)^4p

Therefore the probability that you stop at the fifth flip would be

                                    p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

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