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noname [10]
3 years ago
8

A blueberry recipe calls for 1\frac{3}{4} cups of sugar. How many cups of sugar would need to make the muffin recipe 3 times?

Mathematics
1 answer:
liraira [26]3 years ago
4 0

Answer:

4 3/4

Step-by-step explanation:

Just multiply 1 3/4 and 3 = 4 3/4

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225 quarters are easy to loose and who will accept that many quarters

Step-by-step explanation:

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What expression is equivalent to -3n-7+(-6n)+1
nekit [7.7K]
-3n-7+(-6n)+1 = 0
-3n-7-6n+1 = 0
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Keiko sold 3 less than three-fourths of his sister's sales. Which expression represents what Keiko sold?
lisov135 [29]

Answer: 3/4x - 3

Step-by-step explanation:

Let his sister's sale be represented by x.

Since we are informed that Keiko sold 3 less than three-fourths of his sister's sales, the expression that represents what Keiko sold will be:

= (3/4 × x) - 3

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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.

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There are some cubes in a bag.
dedylja [7]
1:5 is the answer in 1:n form
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