- 2 root 2 + root 2 ?

Which is greater? -4.84 or -8.48

velikii [3]

Answer:

-4.84

Step-by-step explanation:

the -4.84 is closest to the positive side of a number line. meaning -4.84 is greater

4 0

2 years ago

Read 2 more answers

What is 8.04 × 101? 0.804 8.04 8.4 80.4

Lerok [7]

8.04 * 100= 804

You move the decimal point two times to the right

8.04*101=812.04

6 0

2 years ago

Read 2 more answers

In a sample of 1110 U.S. adults, 172 think that most celebrities are good role models. two U.S. adults are selected from this s

marshall27 [118]

Answer: 123

Step-by-step explanation:

5 0

2 years ago

Please helppppppp!!!!!!!!!!!!!!!!!!!!!!!!!!!

Olin [163]

So here, ABCD is being inscribed in a circle, so ABCD is a cyclic quadrilateral (as inscribed in a circle) also, there exists a property of cyclic quadrilateral, i.e, the sum of opposite angles of a cyclic quadrilateral is 180°, so we can just write the following equation using this :

${:\implies \quad \sf y^{\circ}+83^{\circ}=180^{\circ}}$

${:\implies \quad \sf y^{\circ}=180^{\circ}-83^{\circ}}$

${:\implies \quad \sf y^{\circ}=97^{\circ}}$

${:\implies \quad \boxed{\bf{y=97}}}$

6 0

2 years ago

A player of a video game is confronted with a series of four opponents and an 80% probability of defeating each opponent. Assume

mixer [17]

Answer:

(a) 0.4096

(b) 0.64

(c) 0.7942

Step-by-step explanation:

The probability that the player wins is,

$P(W)=0.80$

Then the probability that the player losses is,

$P(L)=1-P(W)=1-0.80=0.20$

The player is playing the video game with 4 different opponents.

It is provided that when the player is defeated by an opponent the game ends.

All the possible ways the player can win is: {L, WL, WWL, WWWL and WWWW)

(a)

The results from all the 4 opponents are independent, i.e. the result of a game played with one opponent is unaffected by the result of the game played with another opponent.

The probability that the player defeats all four opponents in a game is,

P (Player defeats all 4 opponents) = $P(W)\times P(W)\times P(W)\times P(W)=[P(W)]^{4} =(0.80)^{4}=0.4096$

Thus, the probability that the player defeats all four opponents in a game is 0.4096.

(b)

The probability that the player defeats at least two opponents in a game is,

P (Player defeats at least 2) = 1 - P (Player losses the 1st game) - P (Player losses the 2nd game) = $1-P(L)-P(WL)$

$=1-(0.20)-(0.80\times0.20)\\=1-0.20-0.16\\=0.64$

Thus, the probability that the player defeats at least two opponents in a game is 0.64.

(c)

Let X = number of times the player defeats all 4 opponents.

The probability that the player defeats all four opponents in a game is,

P(WWWW) = 0.4096.

Then the random variable $X\sim Bin(n=3, p=0.4096)$

The probability distribution of binomial is:

$P(X=x)={n\choose x}p^{x} (1-p)^{n-x}$

The probability that the player defeats all the 4 opponents at least once is,

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-[{3\choose 0}(0.4096)^{0} (1-0.4096)^{3-0}]\\=1-[1\times1\times (0.5904)^{3}\\=1-0.2058\\=0.7942$

Thus, the probability that the player defeats all the 4 opponents at least once is 0.7942.

3 0

3 years ago