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2 years ago

PLEASE HELP THIS IS WORTH 50 POINTS!!!

Mathematics

2 answers:

alisha [4.7K]2 years ago

8 0

Answer:

You still need the awnser?

Step-by-step explanation:

Pani-rosa [81]2 years ago

3 0

Answer:

Do you still need help

Step-by-step explanation:

You might be interested in

A Rectangular piece of sheet Metal has an area of 10,680 in.² its length is 72.0 in. find it width

tatiyna

148 1/3 inches. 72 x 148 1/3 = 10,680
To get the answer 10,680 / 72 = 148 1/3

5 0

3 years ago

if you think that Andy solved the problem correctly, explain how you can tell he is correct. If not, identify and explain all er

nasty-shy [4]

Answer:

25 lol

Step-by-step explanation:

4 0

2 years ago

Rewrite the rational exponent as a radical by extending the properties of integer exponents. 2 to the 3 over 4 power, all over 2

Zepler [3.9K]

We are asked in this problem to determine the simplified expression of the statement given. The rules that apply in exponential functions is that when an exponential term is raised to the power of an integer, the simplified term has a degree that is equal to the product of the integers involved. The operations involved should be applicable to terms with the same base number only. In this problem, we thus write:

2^3/4 / 2^1/2 = 2^3/4 * 2^-1/2 = = 2^(3/4 - 1/2) = 2^ 1/4. hence the answer is 2^0.25 or simply equal to 1.1892 determined using a calculator.

6 0

2 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

Please answer correctly !!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!!

solong [7]

Answer:

That looks like interior angle to me

x = 47 (since interior angles are congruent)

Step-by-step explanation:

I am really sleepy... :/

8 0

2 years ago