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seropon [69]
3 years ago
14

Leila purchased 21.5 centimeters of wire for $17.20.

Mathematics
1 answer:
Brums [2.3K]3 years ago
4 0

Answer:

17.20 \div 21.5 =

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Answer:

Where’s the question, send it and let me try

Step-by-step explanation:

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If i was born april 16th 1991 how old would i be?
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You would be 26 im pretty sure
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You randomly choose one paper clip from a jar. In the jar there are 6 green, 3 white, 4 red, 2 blue, and 5 yellow paper clips.
Taya2010 [7]

Answer:

3/10

Step-by-step explanation:

Total paper clip =6+3+4+2+5

=20

Probability of green = number of green/total

= 6/20

leaving in it's lowest term

probability of green=3/20

4 0
3 years ago
In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0
4 years ago
*<br> Write the equation in standard form using integers.<br><br> y=-4/5x+3
svetoff [14.1K]

Answer:

4x + 5 y = 15

Step-by-step explanation:

6 0
3 years ago
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