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3 years ago

If mZFBC = (10x – 9), mZCBE = (4x + 15), find mZFBE.

Mathematics

2 answers:

coldgirl [10]3 years ago

6 0

Answer:

mFBE = 14x + 6

Step-by-step explanation:

As you can see FBE is sum of FBC + CBE

Given mFBE = mFBC + mCBE

= 10x - 9 + 4x + 15

= 14x + 6

So mZFBE = 14x + 6

erastovalidia [21]3 years ago

5 0

Answer:

BEFmZ Is the answer i think

Step-by-step explanation:

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Find the solution <br> Y = -3x - 5

igomit [66]

Answer:
-5/3

Explanation:

5 0

2 years ago

let X represent the amount of time till the next student will arriv ein the library partking lot at the university. If we know t

AlekseyPX

Answer:

0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean of 4 minutes

This means that $m = 4, \mu = \frac{1}{4} = 0.25$

Find the probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot:

This is:

$P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2)$

In which

$P(X \leq 132) = 1 - e^{-0.25*132} = 1$

$P(X \leq 2) = 1 - e^{-0.25*2} = 0.393469$

$P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2) = 1 - 0.393469 = 0.606531$

0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.

4 0

3 years ago

What is the sequence of transformations that maps △ABC to △A′B′C′ ? Select from the drop-down menus to correctly identify each s

torisob [31]

<h3>Answer:</h3>

Any 1 of the following transformations will work. There are others that are also possible.

translation up 4 units, followed by rotation CCW by 90°.
rotation CCW by 90°, followed by translation left 4 units.
rotation CCW 90° about the center (-2, -2).

<h3>Step-by-step explanation:</h3>

The order of vertices ABC is clockwise, as is the order of vertices A'B'C'. Thus, if reflection is involved, there are two (or some other even number of) reflections.

The orientation of line CA is to the east. The orientation of line C'A' is to the north, so the figure has been rotated 90° CCW. In general, such rotation can be accomplished by a single transformation about a suitably chosen center. Here, we're told there is <em>a sequence of transformations</em> involved, so a single rotation is probably not of interest.

If we rotate the figure 90° CCW, we find it ends up 4 units east of the final position. So, one possible transformation is 90° CCW + translation left 4 units.

If we rotate the final figure 90° CW, we find it ends up 4 units north of the starting position. So, another possible transformation is translation up 4 units + rotation 90° CCW.

Of course, rotation 90° CCW in either case is the same as rotation 270° CW.

_____

We have described transformations that will work. What we don't know is what is in your drop-down menu lists. There are many other transformations that will also work, so guessing the one you have available is difficult.