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kvasek [131]
3 years ago
10

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

<u>tan(x) = 8.78/12</u>

<u />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?? <br><br>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 <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> 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}

       =0.734

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0
2 years ago
What percent of 37 is 111?
exis [7]
111 is 300% of 37. This is how I do it. 37x=111 x=111/37 x=3 move the decimal point right 2 places to get the percentage 300% ANSWER 300%
4 0
3 years ago
Suppose you buy a CD for $250 that earns 3% and is compounded quarterly. The CD matures in 5 years. Assume that if funds are wit
Eduardwww [97]
The early withdrawal fee on this account
250×0.03×(1÷4)=1.875
8 0
2 years ago
Read 2 more answers
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