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erica [24]
3 years ago
9

The lines p and q intersect at point O. What is the value of x? Enter your answer in the box. x = Lines p and q intersect at poi

nt O. Acute vertical angles measure two x plus thirteen degrees, and three x minus three degrees.
Mathematics
2 answers:
Angelina_Jolie [31]3 years ago
6 0

Answer:

x = 16

Step-by-step explanation

Find the diagram attached. You can see from the diagram that both angles are vertically opposite angles and vertically opposite angles are equal.

Hence 2x+13 = 3x-3

Collect like terms

2x-3x = -3-13

-x = -16

Multiply both sides by -1

-1(-x) = -1(-16)

x = 16

Hence the value of x is 16

Oliga [24]3 years ago
3 0

Answer:

x=16 hope this helps!

Step-by-step explanation:

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In ordinal measurement scale the order of values or variables are very important. Just as in the case above, the rating from 1-3 is according to the order of it's importance.

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7 0
3 years ago
PLS HALPPPPP!!!!
andrey2020 [161]

Answer:

11.01

Step-by-step explanation:

3.78+4.51+2.72=11.01

4 0
3 years ago
Simplify the expression: <br> r -3 -r
Ymorist [56]

Answer:

-3

Step-by-step explanation:

Combine the opposite terms in  r  −  3  −  r .

8 0
3 years ago
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Answer:seventh year

Step-by-step explanation:

4 0
3 years ago
A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin
Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

P(\dfrac{X}{Y}) = 60\% = 0.6

P(\dfrac{X}{Y'}) = 0.01\% = 0.0001

P(Y) = 0.01

Thus, P(Y') \ will\ be = 1 - P(Y)

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

P(YX) = P(\dfrac{X}{Y}) \ P(Y)

P(YX) =0.6 \times 0.01

P(YX) =0.006

The probabilities of not involved in cheating & the evidence are present is:

P(Y'X) = P(Y')  \times P(\dfrac{X}{Y'})

P(Y'X) = 0.99  \times 0.0001 \\ \\  P(Y'X) = 0.000099

(b)

The required probability that the evidence is present is:

P(YX  or Y'X) = 0.006 + 0.000099

P(YX  or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}

P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}

P(\dfrac{Y}{X}) = 0.9838

5 0
3 years ago
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