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djyliett [7]
3 years ago
13

A large scale weigh 3000 pound 5 ounces, which of the following scale weigh in ounces​

Mathematics
1 answer:
Iteru [2.4K]3 years ago
7 0

Answer:

48005 ounces

Step-by-step explanation:

1 pound = 16 ounces

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Part B Find the total number of pounds of cheese ordered. A. 11 pounds B. 13 pounds C. 15 pounds D. 17 pounds
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hi all how are you I am well will you play free fire for participating in give away please make me brainlist give away on 26 4:00 clock send your like see you then

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2 years ago
Mathematics<br>ose the correct answer:<br>. What number should be added to (-5/16) to get ( 7/24)?​
fgiga [73]

Answer:

0.6042 or 29/48

Step-by-step explanation:

-5/16 = -0.3125

7/24 = 0.2917

0.2917 - -0.3125 = 0.6042

0.6042 ≅ 29/48

7 0
3 years ago
Read 2 more answers
In an NBA championship series, the team that wins four games out of seven is the winner. Suppose that teams ???? and ???? face e
olchik [2.2K]

Answer:

a) P=0.185

b) P=0.608

c) P=0.593

Step-by-step explanation:

The question is not complete

<em>"In an NBA championship series, the team that wins four games out of seven is the winner. Suppose that teams A and B face each other in the championship games and that team A has a probability 0.55 of winning a game over team B.</em>

<em>a. What is the probability that team A will win the series in 6 games?</em>

<em>b. What is the probability that team A will win the series?</em>

<em>c. If teams A and B were facing each other in a regional playoff series, which is decided by winning three out of 5 games, what is the probability that team A would win the series? </em>

a) It is implicit that 6 games are played. The number of games needed to win a series in 6 games is 4, and the last winning should be in the 6th game.

So we have 5 random results in the first 5 games in which A team wins 3 matches and losses 2 matches.

This can be modeled as a binomial distribution with k=3, n=5 and p=0.55.

P(k=3)=\frac{n!}{k!(n-k)!}p^k*(1-p)^{n-k}\\\\P(k=3)=\frac{5!}{3!2!}*0.55^3*0.45^2=10*0.166*0.203=0.337

Then, this probability has to be multiplied by the probability of winning the last game.

P(W_6)=P(k=3;n=5)*P(W_{6th})=0.337*0.55=0.185

The probability that team A will win the series in 6 games is P=0.185.

<em />

b) In this case, we have to calculate the probability of A winning at least 4 games. When a team reaches 4 wins, the series is over. This can happen in 4, 5, 6 or 7 games.

The probability of winning the series is:

P(W)=[P(3;3)+P(3;4)+P(3;5)+P(3;6)]*P(win\,last\,game)

Probability of 3 wins in 3 games

P(3;3)=\frac{3!}{3!0!}*0.55^3*0.45^0 =1*0.166*1=0.166

Probability of 3 wins in 4 games

P(3;4)=\frac{4!}{3!1!}*0.55^3*0.45^1 =4*0.166*0.45=0.299

Probability of 3 wins in 5 games

P(3;5)=\frac{5!}{3!2!}*0.55^3*0.45^2 =10*0.166*0.203=0.337

Probability of 3 wins in 6 games

P(3;6)=\frac{6!}{3!3!}*0.55^3*0.45^3 =20*0.166*0.091=0.303

Then

P(W)=[P(3;3)+P(3;4)+P(3;5)+P(3;6)]*P(win\,last\,game)\\\\P(W)=[0.166+0.299+0.337+0.303]*0.55=1.105*0.55=0.608

The probability of team A of winning the series is P=0.608.

c) The probabilty of winning the regional playoff is

P(W)=[P(2;2)+P(2;3)+P(2;4)]*P(win\,last\,game)

P(2;2)=\frac{2!}{2!0!}*0.55^2*0.45^0 =1*0.303*1=0.303\\\\P(2;3)=\frac{3!}{2!1!}*0.55^2*0.45^1 =3*0.303*0.45=0.408\\\\P(2;4)=\frac{4!}{2!2!}*0.55^2*0.45^2 =6*0.303*0.203=0.368

Then,

P(W)=[P(2;2)+P(2;3)+P(2;4)]*P(win\,last\,game)\\\\P(W)=[0.303+0.408+0.368]*0.55=1.079*0.55=0.593

8 0
2 years ago
A: bag a <br> B :bag b<br> C :bag c<br> D : bag d
Luden [163]

Answer:

Your answer is <em>Bag B</em>

<em>Hope this helped my love!</em>

<em>Can i please have the brainliest?</em>

6 0
3 years ago
Urgent please solve quickly please!
loris [4]

Answer: x = 0, y = 7

-x+y=7

add x to both sides

y=x+7

plug this in for y in the other equation

6x-(x+7)=-7​

simplify

6x-x-7=-7

simplify

5x-7=-7

add 7 to both sides

5x=0

divide both sides by 5

x=0

plug in to one of the equations to find y

-(0)+y=7

y=7

Therefore, x=0 and y=7.

Hope this helps!

6 0
3 years ago
Read 2 more answers
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