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2 years ago

Context is supported below so i need to know what the value of C(0) is

Mathematics

2 answers:

AlekseyPX2 years ago

8 0

The answer is 5838390 because

Mars2501 [29]2 years ago

6 0

Answer:

ill answer in a minute

Step-by-step explanation:

im going to do it all its going to take a sec

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Solve for x:1/2(x-2)+1/3(x-3)+1/4(x-4)

Archy [21]

Answer:

$\frac{1}{12} \times (13x - 36)$

4 0

2 years ago

Meghan's favorite music group has 80 songs. Meghan has 75% of the

kherson [118]

Answer:

60 songs

Step-by-step explanation:

6 0

2 years ago

Captain Salamander has a lazy time on the trip and spends about 15 hours a day sleeping. What fraction of the day does he spend

padilas [110]

Well there is 24 hours in a day so 24 - 15 = 9. So he spends 9 hours awake.

7 0

2 years ago

HEY CAN ANYONE PLS ANSWER DIS MATH QUESTION!!!!!!!!

ziro4ka [17]

Answer:

80 tons
$20,000

Step-by-step explanation:

<u>21. 1st scenario</u>

B = fy + wz
100,000 = 15,000*4 + 500w
500w = 100,000 - 60,000
500w = 40,000
w = 40,000/500
w = 80

80 tons of wheat

<u>22. 2nd scenario</u>

B = fy + wz
50,000 = 2y + 200*50
50,000 = 2y + 10,000
2y = 50,000 - 10,000
2y = 40,000
y = 20,000

Fighter plane's price is $20,000

8 0

3 years ago

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.

7 0

2 years ago