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MariettaO [177]
3 years ago
8

If the given values in a right triangle are cos A =

Mathematics
1 answer:
Vlad1618 [11]3 years ago
5 0
See the picture attached to better understand the problem

we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²---->  equation 1

applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2

substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2

the answer is
the hypotenuse is (3√2)/2

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cestrela7 [59]

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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

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