Pls help me with 15-19 pls

Explain why ΔABC cannot be shown to be congruent to ΔCDA.

igomit [66]

There isn't enough info to prove the triangles to be congruent or not. So we can't say for sure either way.

We have angle CAD = angle ACB given by the arc markings, and we know that AC = AC due to the reflexive theorem. However we are missing one third piece of information.

That third piece of info could be....

AD = BC which allows us to use SAS
angle ACD = angle CAB which allows us to use ASA
angle ABC = angle CDA which allows us to use AAS (slight variation of ASA)

Since we don't know any of those three facts, we simply don't have enough information.

side note: If AB = CD, then this leads to SSA which is not a valid congruence theorem. If we had two congruent sides, the angle must be between the two sides, which is what AD = BC allows.

4 0

2 years ago

Find the probability of each event. There are 4 red marbles, 2 black marbles, and 2 green marbles in a box. P(red) = P ( red or

nevsk [136]

Answer: P(red): 1/2

P(red or black):3/4

Step-by-step explanation: Count the total number of marbles. There are 8. Since there are 4 red marbles out of the 8, put this into a fraction. It will be 4/8, but simplify to 1/2. The P(red) is 1/2.

Since there are 4 red marbles and 2 black marbles, add them. There are 6 out of the 8 marbles. The fraction is 6/8 simplify to 3/4. The P(red or black) is 3/4.

7 0

3 years ago

6. Assume that a component passes a test is 0.85 and that components perform independently. What is the probability that the thi

Tanzania [10]

Answer:

3.90% probability that the third failure will occur on the tenth component tested

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it fails, or it does not fail. Components perform independently. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Assume that a component passes a test is 0.85

So they fail with probability of $p = 1 - 0.85 = 0.15$

What is the probability that the third failure will occur on the tenth component tested

First 9 components: Two failures, that is, P(X = 2) when n = 9.

10th component: Failure with probability 0.15.

So

$P = 0.15P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{9,2}.(0.15)^{2}.(0.85)^{7} = 0.2597$

So

$P = 0.15P(X = 2) = 0.15*0.2597 = 0.0390$

3.90% probability that the third failure will occur on the tenth component tested

5 0

3 years ago

Which of the following is an example of a proper fraction?

nirvana33 [79]

The numbers should be less than the denominator, therefore i think it's 5/9 , because the rest have a numerator greater than the denominator, even the 3 . it's written as 3/1 and 3 >1.

4 0

3 years ago

If a=8 and b=3, evaluate the following expression:2a+3b

hjlf

Answer:

25

Step-by-step explanation:

okay, so we plug our numbers into the equation:

2a+3b = 2x8+3x3

2x8 = 16

3x3=9

16+9 = 25

7 0

3 years ago

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Pls help me with 15-19 pls​

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