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mash [69]
3 years ago
12

I have a 3 foot board that needs to be equally decided into 1/2 how many extra peices will I have

Mathematics
1 answer:
anygoal [31]3 years ago
7 0

Answer:

No extra pieces.

Step-by-step explanation:

Given:

Total number of foot board = 3

Equally divided into =  1/2

Question asked:

How many extra pieces will I have = ?

Solution:

By dividing 3 by 1/2 we can find the extra pieces.

                                                                 = 3\div\frac{1}{2}

                                                                 = 3\times\frac{2}{1}

                                                                 = 6

Since, the remainder is zero, thus there is no any extra pieces I have.

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Tamara and Amir shared a candy bar. Tamara ate two fifths. Amir ate two fifths. How much is left?
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Answer:

1/5

Step-by-step explanation:

Both Amir and Tamara ate 2/5 each.

2/5 + 2/5 = 4/5

If 4/5 of the candy bar was eaten, 1/5 of the candy bar remains because

5/5 - 4/5 = 1/5

(Remember that 5/5 just equals one whole, representing the whole candy bar)

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3 years ago
Which of the following shows the solution to the inequality
otez555 [7]

Answer:

I got x = 1

Step-by-step explanation:

First to isolate the variable you add 3 to both sides which will give you 1/2x is less than or equal to 2. Then divide both sides by 1/2 and you will get x = 1. To check the answer I just did:    

1/2(1) -3 \leq -1\\1/2 -3\leq -1\\-2.5\leq -1

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7 0
2 years ago
0.5% of 490 is what number  show work please
Genrish500 [490]
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6 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
3 years ago
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