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Vedmedyk [2.9K]
3 years ago
9

A bag holds 6 tiles: 2 lettered and 4 numbered. Without looking, you choose a tile. What is the probability of drawing a number?

Mathematics
2 answers:
sergiy2304 [10]3 years ago
5 0
You take the total number as the denominator and the amount of each become your numerators.For numbered it's 4/6 then you divide both the top and bottom number by two and you get 2/3 or 66%.
e-lub [12.9K]3 years ago
4 0
4/6 because the bag holds six tiles total so 4 of the 6 are numbered so you would have 4/6 of a chance of picking a numbered tile.
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Stolb23 [73]
(-6.5,0)

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7 0
3 years ago
Select the correct answer from each drop-down menu.
sdas [7]

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3 0
2 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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larisa [96]
Slope point form :
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Y - y1 = m(X-X1)

But before, we must solve for the m value or slope.
M = y2-y1/x2-X1
M = 5/2 - -1/2 / -1/2 - 3/2.
M = 5/2 + 1/2 / -(1/2+3/2)
M = 6/2 / -(4/2)
M = 3/-2

Now we can place it in slope point and also in standard form of a line.

Y-y1 = m(X -X1)

Y - -1/2 = -3/2(X - 3/2)

Y + 1/2 = -3/2(X - 3/2)

This is in slope point form.

Y + 1/2 = -3/2x + 9/4
Y + 1/2 - 1/2 = -3/2x + 9/4 - 1/2
1/2 = 2/4
Y = -3/2x + 7/4
-3/2x = -6/4x
Y = -6/4x + 7/4
Y • 4 = 4( -6/4 X + 7/4)
4y = -6x + 7
4y + 6x = -6x + 6x +7

6x + 4y = 7

This is in standard form. If you have any questions of the steps just ask.
4 0
3 years ago
How do you write 7258630 in word form
Aloiza [94]
7,258,630-

seven million, two hundred fifty-eight thousand, six hundred thirty :-)
8 0
3 years ago
Read 2 more answers
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