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Cerrena [4.2K]
3 years ago
13

Bill owed his brother $21.50, but was able to pay him back $18.75. How much does Bill owe his brother now?

Mathematics
1 answer:
Andrew [12]3 years ago
5 0

Answer:

2.75

Step-by-step explanation:

21.50 - 18.75 = $2.75

CAN I HAVe BRAINLIEST

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which expression represents the amount of money john has collected? john collects $5 for club dues from each member.
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5.00x (number of people)
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What is the remainder of x^5+2x^4+9x^3-6x^2+3x+3165 divided by x-5
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Answer:

8530

Step-by-step explanation:

The remainder is 5⁵ + 2(5)⁴ + 9(5)³ - 6(5)² + 3(5) + 3165 = 8530

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The total charge on 6 particles is -48units. All the particles have the same charge. What is the charge on each particle?
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i believe the answer will be -7

4 0
3 years ago
Let x be the amount of time (in minutes) that a particular San Francisco commuter must wait for a BART train. Suppose that the d
larisa [96]

Answer:

a) P(X

P(X>14) = 1-P(X

b) P(7< X

c) We want to find a value c who satisfy this condition:

P(x

And using the cumulative distribution function we have this:

P(x

And solving for c we got:

c = 20*0.9 = 18

Step-by-step explanation:

For this case we define the random variable X as he amount of time (in minutes) that a particular San Francisco commuter must wait for a BART train, and we know that the distribution for X is given by:

X \sim Unif (a=0, b =20)

Part a

We want this probability:

P(X

And for this case we can use the cumulative distribution function given by:

F(x) = \frac{x-a}{b-a} = \frac{x-0}{20-0}= \frac{x}{20}

And using the cumulative distribution function we got:

P(X

For the probability P(X>14) if we use the cumulative distribution function and the complement rule we got:

P(X>14) = 1-P(X

Part b

We want this probability:

P(7< X

And using the cdf we got:

P(7< X

Part c

We want to find a value c who satisfy this condition:

P(x

And using the cumulative distribution function we have this:

P(x

And solving for c we got:

c = 20*0.9 = 18

3 0
3 years ago
In the set of consecutive integers from 15 to 30 inclusive, two integers are multiples of both 3 and 5. How many integers in thi
Lisa [10]

Answer:

Step-by-step explanation:

To make it easy let's list the integers that are beetwen 15 to 30 inclusive :

15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30

  • the numbers that are multiples of both 5 and 3 are ending with 5 and 0 and the sum o their digits is a multiple of 3
  • 15,30 oly the values

the the whole interval ]15,30[ is the answer

8 0
3 years ago
Read 2 more answers
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