All categories

2 years ago

Which of the following can be used to show that these triangles are congruent?

Mathematics

1 answer:

Illusion [34]2 years ago

8 0

A
The hidden information is that angleACB is equal to angle DCE
Brainliest pls!

You might be interested in

Help ASAP Please check to see if I have the last part correct. Thanks

7nadin3 [17]

<span>In 10 repetitions, there was an instance where 3 or more out of 5 free throw missed. So the probability of her missing 3 or more out of 5 free throw is 0/10=0% or 0.0%. I</span>t appears that your answer of 0% is correct given your experiment.

3 0

3 years ago

Read 2 more answers

260 years equals:

uranmaximum [27]

It would have to be letter D my guy, if you know what centuries and decades are this is easy

6 0

3 years ago

Read 2 more answers

What's the vertex for 6x^2-48x-54=0?

kakasveta [241]

It would be 4,-25 becuase of in the first place i in and the other imp in cams was the

4 0

2 years ago

WILL MARK AS BRAINIEST!!!!!!! In this triangle, which of the following is true?

prisoha [69]

Answer:

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°

Subtract the sum of the 2 given angles from 180

x = 180° - (90 + 35)° = 180° - 125° = 55°

----------------------------------------------------------

Since the triangle is right use the cosine ratio to solve for b

cos35° = $\frac{adjacent}{hypotenuse}$ = $\frac{b}{20}$

Multiply both sides by 20

20 × cos35° = b, hence

b = 16.38 ( to 2 dec. places )

7 0

3 years ago

Read 2 more answers

A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0

yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{10-x}$

Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

$P(x$

Where, for example, P(0) and P(1) are calculated as:

$P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098$

Then, $P(x$ , so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

$P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)$

Where, for example, P(4) is calculated as:

$P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001$

Then, $P(x\geq4 )=0.3504$ , so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

7 0

3 years ago