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2 years ago

What is the equation of the circle with center (0, 0) that passes through the point (–4, –6)?

Mathematics

1 answer:

Vadim26 [7]2 years ago

4 0

If the center is in point P(a,b) and has a radius r then circle equation is:

(x-a)²+(y-b)²=r².

In your task we've got P(0,0) so the circle equation is x²+y²=r².

We have to find r. Radius start at (0,0) and end at (-4,-6) therefore

r=√((-4)²+(-6)²) and

r²=(-4)²+(-6)²=16+36=52.

So answer is x²+y²=52.

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Let B ∼ Bin(20,0.2). Compute the following probabilities. I would suggest computing these with a hand calculator using the formu

Makovka662 [10]

Answer:

a) 0.2182

b) 0.0691

c) 0.9309

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly b successes on n repeated trials, and B can only have two outcomes.

$P(B = b) = C_{n,b}.p^{b}.(1-p)^{n-b}$

In which $C_{n,b}$ is the number of different combinations of b objects from a set of n elements, given by the following formula.

$C_{n,b} = \frac{n!}{x!(n-b)!}$

And p is the probability of B happening.

In this problem we have that:

Bin(20,0.2).

This means that $n = 20, p = 0.2$

(a) P(B=4).

$P(B = b) = C_{n,b}.p^{b}.(1-p)^{n-b}$

$P(B = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

(b) P(B≤1).

$P(B \leq 1) = P(B = 0) + P(B = 1)$

$P(B = b) = C_{n,b}.p^{b}.(1-p)^{n-b}$

$P(B = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(B = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(B \leq 1) = P(B = 0) + P(B = 1) = 0.0115 + 0.0576 = 0.0691$

(c) P(B>1).

Either B is less than or equal to 1, or B is larger than 1. The sum of the probabilities of these events is decimal 1. So

$P(B \leq 1) + P(B > 1) = 1$

We have that, from b), $P(B \leq 1) = 0.0691$

$0.0691 + P(B > 1) = 1$

$P(B > 1) = 0.9309$

8 0

2 years ago

Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?

irakobra [83]

First, note that $m\angle A+m\angle C=90^{\circ}.$ Then

$m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.$

Consider all options:

$\tan A=\dfrac{\sin A}{\sin C}$

By the definition,

$\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.$

Now

$\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.$

Option A is true.

$\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.$

Then

$\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.$

Option B is false.

$\sin C = \dfrac{\cos A}{\tan C}.$

By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

Now

$\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.$

Option C is false.

$\cos A=\tan C.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

As you can see $\cos A\neq \tan C$ and option D is not true.

$\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.$

By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.$

Then

$\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.$

This option is false.

8 0

2 years ago

Read 2 more answers

Reza paid $4,500 as a down payment on a car. She then made equal monthly paments of $250. Reza paid a total of $10,500 for her c

monitta

10500-4500=6000

6000/250=24

So the answer is 24 months, or 2 years.

6 0

2 years ago

Scores on an exam are normally distributed with a mean of 76 and a standard deviation of 10. In a group of 230 tests, how many s

Tanzania [10]

The score of 96 is 2 standard deviations above the mean score. Using the empirical rule for a normal distribution, the probability of a score above 96 is 0.0235.
Therefore the number of students scoring above 96 is given by:
$230\times 0.0235=5\ students$

5 0

2 years ago

Read 2 more answers

How much would Carol have to invest today at 6.2% compounded annually to have $4600 for a vacation to China in two years?

KiRa [710]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$4600\\
P=\textit{original amount deposited}\\
r=rate\to 6.2\%\to \frac{6.2}{100}\to &0.062\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, so once}
\end{array}\to &1\\

t=years\to &2
\end{cases}
\\\\\\
4600=P\left(1+\frac{0.062}{1}\right)^{1\cdot 2}$

solve for P

5 0

2 years ago