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3 years ago

If BD and EG are parallel lines and mBCF = 50°, what is mEFH? Look at this diagram:

Mathematics

1 answer:

Burka [1]3 years ago

4 0

It’s hard to tell due to no labels, but these are your angles due to the equal angles theorem and a few other theorems. Hopefully with this image you can hopefully find your answer. You may need to add angles together depending on what you are looking for. And possible even figure out what angle measure would make a (180 degree) triangle.
Hope this helped

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How do functions work? Could someone explain how to know which questions are functions or not

belka [17]

Answer:

A function is an equation that has only one answer for y for every x. A function assigns exactly one output to each input of a speceified type. It is common to name a function either f(x) or g(x) instead of y. f(2) means that we should find the value of our function when x equals 2.

Step-by-step explanation:

here is a step by step equation:

f(x) = x + 7

if:

x=2

then:

f(2) = 2 + 7 = 9

A function is linear if it can be defined by

f(x) = mx + b

Makes sense? Hope I helped

plus I'm in middle school learning this also.

6 0

3 years ago

How do I delete this

elena-s [515]

Answer:

I hope this helps!

Step-by-step explanation:

Have a great rest of your day!

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3 years ago

A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the fir

erica [24]

Answer:

(a) 0.7

(b) $\frac 13$

Step-by-step explanation:

Let $E_1, E_2,$ and $E_3$ be the events of passing the 1st, 2nd, and 3rd exam individually.

Given that $P(E_1)=0.9\;\cdots(i)$

where $P(E_1)$ denotes the probability of passing the 1st exam.

Condition for passing the second exam is, at first, the candidate must have to pass the 1st exam.

So, $P\left(\frac{E_2}{E_1}\right)=0.8\;\cdots(ii)$

where $P(E_2/E_1)$ denotes the probability of passing the 2nd exam when she already passed the 1st exam (given, 0.8).

Similarly, as the conditional probability of passing the 3rd exam is 0.7, and the condition for this is, at first, she must have to pass the 1st and 2nd exam. i.e,

$P\left(\frac{E_2}{P(E_2/E_1)}\right)=0.7\;\cdots(iii)$

(a) For passing all the exams, the condition is, at first, she has to pass the 1st and 2nd exam, then she has to pass the 3rd exam too. The probability for this conditional has been given as 0.7.

So, the probability that she passes all three exams is 0.7.

(b) Given that she didn't pass all three exams that means she either failed in 1st exam or she passed the 1st and failed in 2nd exam or she passed both 1st and 2nd but failed in the 3rd exam.

Let F be the event that she didn't pass all three exams. So,

$P(F)=(1-P(E_1))+\left(1-\frac{P(E_2)}{P(E_1)}\right)+\left(1-\frac{P(E_2)}{P(E_2/E_1)}}\right)$