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erica [24]
2 years ago
7

A game has a rectangular board with an area of 44in2. There is a square hole near the top of the game board in which you must no

t toss in a bean bag. The square has side lengths of 3in. What is the probability of tossing the bag through the hole twice in a row?
9/44
81/1936
1225/1936
35/44
Mathematics
1 answer:
Evgen [1.6K]2 years ago
4 0
There's a key piece of information that we have to assume about this problem: that there is a 100% probability of hitting the game board. (you will definitely hit the board). That means the probability of getting into the hole is the area of the hole divided by the area of the board. 

The area of the board is given: 44 in²
The area of the hole is 3x3 = 9 in²

The probability of hitting the hole on any given try:
9/44 
Twice in a row:
9/44 * 9/44 = 81/1936

Answer is B) 81/1936 
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LenaWriter [7]

Answer:

The probability that the pirate misses the captain's ship but the captain hits = 0.514

Step-by-step explanation:

Let A be the event that the captain hits the pirate ship

The probability of the captain hitting the pirate ship, P(A) = 3/5

Let B be the event that the pirate hits the captain's ship

The probability of the pirate hitting the captain's ship P(B) = 1/7

The probability of the pirate missing the captain's ship, P'(B) = 1 - P(B)

P'(B) = 1 - 1/7 = 6/7

The probability that the pirate misses the captain's ship but the captain hits = P(A) * P(B) = 3/5 * 6/7

= 0.514

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2 years ago
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A small ferryboat is 4.00 m wide and 6.00 m long. When a loaded truck pulls onto it, the boat sinks an additional 4.00 cm into t
Svet_ta [14]

Answer:

F_w=9408\ N

Weight of the truck=9408 N

Step-by-step explanation:

Boat is experiencing the buoyant force as it is in the water and is sinking

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F_b-F_w=0

where:

F_b is the buoyant force

F_w is the weight=mg

F_b=F_w        Eq (1)

Buoyant force is equal to the mass of water displaced * gravitational acceleration.

F_b=m_{water\ displaced}*g\\F_b=\rho Vg\\

Taking density of water to be 1000 Kg/m^3

F_b=1000*(6*4*4*10^{-2})*9.8\\F_b=9408 N

From Eq(1):

F_w=9408\ N

Weight of the truck=9408 N

6 0
2 years ago
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sp2606 [1]
The answer is
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3 0
2 years ago
Suppose that in a survey of 1,000 U.S. residents, 721 residents believed that the amount of violent television programming had i
Art [367]

Answer: 0.813

Step-by-step explanation:

Let A be the event describes the number of residents believed that the amount of violent television programming had increased over the past 10 years.

& B be the event describes the number of residents believed that the amount of violent television programming had decreased over the past 10 years.

Given : n(A)=721   ;  n(B)=454  ;  n(A∩B)=362

We know that ,

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i.e. n(A\cup B)=721+454-362=813

Also, the total number of U.S. residents surveyed n(S)= 1,000

Then, the  proportion of the 1,000 U.S. residents believed that either the amount of violent programming had increased or the overall quality of programming had decreased over the past 10 years will be :-

P(A\cup B)=\dfrac{n(A\cup B)}{n(S)}=\dfrac{813}{1000}=0.813

Hence, the required answer = 0.813

3 0
2 years ago
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