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Finger [1]
3 years ago
11

2.) What value for c will make the expression a perfect square trinomial?

Mathematics
2 answers:
andre [41]3 years ago
6 0

Answer : \frac{49}{4}

Given expression is x^2 - 7x + c

To make perfect square trinominal we use completing the square method

In completing the square method we add and subtract the half of square of coefficient of middle term

Here coefficient of middle term is -7

Half of -7 is \frac{-7}{2}

Square of \frac{-7}{2} is (\frac{-7}{2})^2 = \frac{49}{4}

So the expression becomes x^2 - 7x +  [tex] \frac{49}{4} that gives perfect square trinomial

Hence , the value of 'c' is \frac{49}{4}


kirill115 [55]3 years ago
6 0
49/4 is the answer :)))
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Step-by-step explanation:

144/60 = 2.4

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CaHeK987 [17]

Answer:B

Step-by-step explanation:

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8 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
What is the value of n in the equation 1/2 (n-4)-3=3-(2n+3)?
patriot [66]

Answer:

Solve for

n

by simplifying both sides of the equation, then isolating the variable.

n

=

2

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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