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levacccp [35]
2 years ago
12

Find the length of segment LM.

Mathematics
2 answers:
-BARSIC- [3]2 years ago
6 0

Answer:

5 I think l to n is 10,so it will be half of it

castortr0y [4]2 years ago
5 0
Segment LM would be 7
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The standard form of the equation of a circle is (x−5)2+(y−6)2=1.
Crank

Answer:

(A)x^2+y^2-10x-12y+60=0

Step-by-step explanation:

The standard form of the equation of a circle is given as:

(x-5)^2+(y-6)^2=1

Simplifying the above given equation, we get

x^2+25-10x+y^2+36-12y=1

x^2+y^2-10x-12y+36+25=1

x^2+y^2-10x-12y+61=1

x^2+y^2-10x-12y+60=0

which is the required general form of the equation.

Hence, option A is correct.

3 0
3 years ago
The 2nd term of an exponential sequence is 9 while the 4th term is 81.find the common ratio,the first term and the sum of the fi
Thepotemich [5.8K]

Answer:

second term: 9

4th term:81

(3rd \: term)^{2}  = 9 \times 81

=729

\sqrt{729}  = 27

3rd term=27

{9}^{2}  = a1x27 \\ 81 = 27a1

a1=3 the first term

81 = 3 \times {q}^{3}

{q}^{3}  = 27

q=3

s = 3x \frac{1 -  {3}^{5} }{1 - 3}  = 364.5

5 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
Which statement is true regarding the graphed functions?
Novosadov [1.4K]
I’m not exactly sure but maybe f(0) = 4 and g(–2) = 4
4 0
2 years ago
50 POINTS ANSWER FAST PLEASE
Sholpan [36]

Answer:

Question 2.) 16

144÷9=16

Question 3.) 16

16÷12=1.33... and all the others equal 1.33... as well

Question 4.) 20

20÷16=1.25   and all tho others equal 1.25 as well

Question 5.)12

8÷12=0.66...    and all the others equal 0.66.... as well

I hope this is good enough for you:

6 0
2 years ago
Read 2 more answers
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