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Xelga [282]
2 years ago
8

Help!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Lilit [14]2 years ago
7 0

Answer:

$5.80

Step-by-step explanation:

butterscotch + milk

butterscotch:

$1/5 = $0.20 x 4 = $0.80

milk:

$1 x 5 = $5

$5 + $0.80 = $5.80

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4 is to 5 as 10 is to a<br><br><br> A.8<br> B.12<br> C.12.5<br> D.20
nexus9112 [7]
Use a proportion.

4 is to 5 as 10 is to a

4/5 = 10/a

4a = 5 * 10

4a = 50

a = 12.5

Answer: C. 12.5
6 0
3 years ago
What is 3 5/8 minus 2.5
Allushta [10]

Answer:

1.125

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
HELP PLEASE, WILL MARK BRAINLIEST
sasho [114]

Answer:

a

Step-by-step explanation:

b square d -for ad +or -2a

4 0
3 years ago
The sum of 5 consecutive integers is 120. What is the third number in the sequence?
Katena32 [7]
5 consecutive integers can be written like
1. x
2. x+1
3. x+2
4. x+3
5. x+4

sum means add them all up.

x  + x + 1 + x + 2 + x + 3 + x + 4 = 120
combine like terms and solve for x.

remember it is looking for third in the sequence, so add 2 to your answer.

good luck!
6 0
3 years ago
Alice searches for her term paper in her filing cabinet, which has several drawers. She knows thatshe left her term paper in dra
katen-ka-za [31]

You made a mistake with the probability p_{j}, which should be p_{i} in the last expression, so to be clear I will state the expression again.

So we want to solve the following:

Conditioned on this event, show that the probability that her paper is in drawer j, is given by:

(1) \frac{p_{j} }{1-d_{i}p_{i}  } , if j \neq i, and

(2) \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i}  } , if j = i.

so we can say:

A is the event that you search drawer i and find nothing,

B is the event that you search drawer i and find the paper,

C_{k}  is the event that the paper is in drawer k, k = 1, ..., n.

this gives us:

P(B) = P(B \cap C_{i} ) = P(C_{i})P(B | C_{i} ) = d_{i} p_{i}

P(A) = 1 - P(B) = 1 - d_{i} p_{i}

Solution to Part (1):

if j \neq i, then P(A \cap C_{j} ) = P(C_{j} ),

this means that

P(C_{j} |A) = \frac{P(A \cap C_{j})}{P(A)}  = \frac{P(C_{j} )}{P(A)}  = \frac{p_{j} }{1-d_{i}p_{i}  }

as needed so part one is solved.

Solution to Part(2):

so we have now that if j = i, we get that:

P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}

remember that:

P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}

this implies that:

P(A \cap C_{j}) = P(C_{j}) \cdot P(A|C_{j}) = p_{i} (1-d_{i} )

so we just need to combine the above relations to get:

P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i}  }

as needed so part two is solved.

8 0
4 years ago
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