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3 years ago

I need help! would appreciate it

Mathematics

2 answers:

pashok25 [27]3 years ago

8 0

Give the person above brainly
I only answered so they could get brainly
Good luck

Ad libitum [116K]3 years ago

5 0

Answer:

x = 0.6

Step-by-step explanation:

We use tangent because the given numbers are opposite and adjacent from x

So, tangent (x) = opposite/adjacent

tan(x) = 8.78/12

$tan^{-1}(\frac{8.78}{12}) = x$

x = 0.631664

Round to nearest tenth:

x ~ 0.6

You might be interested in

Which of the following infinite series has a finite sum? Explain your reasoning.

ikadub [295]

Answer:

d. 20+10+5+2.5+...

Step-by-step explanation:

No geometric series with a common ratio of magnitude greater than 1 will have a finite sum. Nor will any arithmetic series.

Those descriptions exclude answer choices a, b, c. Choice d is a geometric series with a common ratio of 1/2, so will have a finite sum. (It is 40.)

7 0

2 years ago

Help please?? Answer:

Sveta_85 [38]

1. Yes, a regular hexagon can be drawn using rotations.
2.To find the answer, first find the number of sides of a hexagon. A hexagon has six sides.
Divide 360 by 6 = 360/6=60 degrees.
So the angle of rotation for a point on the circle for drawing a regular hexagon is 60 degrees.

3 0

2 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$