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NikAS [45]
3 years ago
6

Which point could represent 5/3? B -3A -1/2D 1/2C 1 1/2​

Mathematics
2 answers:
N76 [4]3 years ago
5 0

Answer:

C. 1 1/2

Step-by-step explanation:

aliya0001 [1]3 years ago
4 0

Answer:

(D)

Step-by-step explanation:

is the correct because is 1/2

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Is this a right triangle?
vesna_86 [32]

Answer:

yes!

Step-by-step explanation:

if it has a 90° angle it is a right triangle :)

8 0
3 years ago
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Which expression is equivalent to (q^4)^-3/q^-15 for all values of q where the expression is defined?
vodka [1.7K]

Answer: (h)q^3

Step-by-step explanation:

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
What is equivalent to 11 - ( - 3 5/8)?
fredd [130]

Step-by-step explanation:

11 - ( - 3 5/8) will equivalent to

11 + 3⅝ = 14⅝

4 0
3 years ago
Solve the following: 3.12 × 1,000 = _____.
torisob [31]

Answer:

60000000.00

Step-by-step explanation:

hgfgjffhvffjfujcdhbcfvjnbcccdsd

8 0
3 years ago
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