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3 years ago

Please help, its so simple

Mathematics

2 answers:

wariber [46]3 years ago

8 0

Answer:

117 degrees

Step-by-step explanation:

Here, we need to use the fact that there is 180 degrees in each triangle. Let's first find the angle that is supplementary to the angle labelled 101 degrees.

Because these two angles are supplementary, they add up to 180, so we can find that interior angle by subtracting 101 from 180: 180 - 101 = 79 degrees.

Now, we know two angles in the triangle: 38 degrees and the one we just found, 79 degrees. To find the third angle, we just subtract 38 and 79 from 180 (because all angles in a triangle add up to 180): 180 - 38 - 79 = 63 degrees.

Notice that this angle that is 63 degrees is actually the supplementary angle if x, so simply subtract 63 from 180 to get x: 180 - 63 = 117 degrees.

Hope this helps!

antoniya [11.8K]3 years ago

5 0

Answer:

117

Step-by-step explanation:

Subtract from 180 to get interior and exterior angles.

180 - 101 = 79

79 + 38 = 117

180 - 117 = 63

180 - 63 = 117

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Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

Step-by-step explanation:

Denote the events as follows:

D = Dave will show up.

M = Mike will show up.

Given:

$P(D^{c})=0.55\\P(M^{c})=0.45$

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

$P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25$

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

$=1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75$

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

$P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25$

Thus, the probability that neither Dave nor Mike will show up is 0.25.

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