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3 years ago

Can someone give me the answer plz.

Mathematics

1 answer:

Solnce55 [7]3 years ago

8 0

Answer : 0.3 as it’s 6/20 :)

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Write an expression that represents the volume of a rectangular prism (box) with length 2003-02-02-00-00_files/i0260000.jpg, wid

k0ka [10]

h=5 in

w-6 in

l=12 in

SA/V=2*12*6+2*6*5+2*12*5/12*6*5

817:180

This is just an example do not use this exact equation and number! Hope it helps. : )

3 0

3 years ago

Openstudt what are the coordinates of the turning point for the function f(x) = (x − 2)3 + 1? (−2, −1) (−2, 1) (2, −1) (2, 1)

Dominik [7]

f(x) = (x - 2)^3 + 1

Find the derivative:-

f'(x) = 3(x -2)^2 This = 0 at the turning points:-

so 3(x - 2)^2 =

giving x = 2 . When x = 2 f(x) = 3(2-2)^3 + 1 = 1

Answer is (2, 1)

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3 years ago

Which mathematical property is demonstrated?

Olin [163]

Answer:

I believe the answer would be A

Step-by-step explanation:

Hope this helps!

5 0

3 years ago

Read 2 more answers

136 X 24 in Standard Algorithm

Ipatiy [6.2K]

3264 and 16,415

Step-by-step explanation:

136 x 24= 3264

245 x 67 = 16,415

4 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

3 years ago