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snow_lady [41]
3 years ago
11

Someone purchase 12 books. But needs 4 books for each of the projects ar school. How many projects would they need

Mathematics
1 answer:
Fudgin [204]3 years ago
7 0
3 projects.
12/4=3. Please mark me brainliest rly wanna mark up!
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There is 4 students in the class 3 ran away how many are left?
pochemuha

Answer:

1 student

Step-by-step explanation:

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2 years ago
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A children's slide has a length of 5.8m. The vertical ladder is 2.6m above the groubd. Find the angle the slide makes with the g
Elodia [21]

<u>Answer</u>


26.6°


<u>Explanation</u>

You are required to use the trigonometric ratio, sine, to find that angle.

SinФ = opposite/hypotenuse


sin Θ = 2.6/5.8

= 0.4483


The angle Θ = Anti-sine(0.4483)

= 26.633°


The angle the slide makes with the ground, correct to one decimal place is 26.6°

8 0
3 years ago
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Cos Z = 32/40 => Z= cos-¹(32/40) =>

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4 0
2 years ago
Each question is asking something different please round will mark brainliest
11111nata11111 [884]

Answer:

1) 78.54

2) 490.87

3) diameter: 15.9999 so 16.00? circumference: 50.27

Step-by-step explanation:

3 0
3 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 <em>X</em> = 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 (<em>X</em> ≥ 1) = 1 - P (<em>X</em> < 1)

             = 1 - P (<em>X</em> = 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
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