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3 years ago

13

Someone help out with this problem

1 answer:

zzz [600]3 years ago

5 0

Answer:

copying angle C

Step-by-step explanation:

it shows that angle c is being copied

hope it helps

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Why would 6% as a fraction be 6/100?

dangina [55]

Well, percent (%) means per hundred, and your number is 6. Therefore, you have 6 per hundred, or 6/100. Also, when you have a percent and want to convert it into a fraction, you must divide the percent by 100, so you could just put the six over 100 since the line between the numerator and denominator basically means division.

3 0

3 years ago

Line k is parallel to line l. Lines k and l are parallel. Lines m and n intersect to form 2 triangles. The top triangle has angl

larisa86 [58]

Answer:The line that is congruent to angle 4 is angle 1

hope this helped :)

6 0

3 years ago

Read 2 more answers

Please helpppppp meeeee!

yan [13]

B. 1 pint
2 cups = 1 pint
4 cups = 1 quart
16 cups = 1 gallon
hope this helps!

7 0

2 years ago

Read 2 more answers

Helppppppp pleaseeee it’s mathhhhhh

Anuta_ua [19.1K]

Answer:

73

Step-by-step explanation:

6x+25=10x-11

-4x=-36

x=8

6(8)+25

48+25=73

I am not sure on this one but hope I helped

3 0

2 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