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3 years ago

Solve the system by graphing. Select the solution(s). y = x + 2 y = 0.5(x + 2)

Mathematics

1 answer:

faltersainse [42]3 years ago

4 0

Answer:

Part 1) The solution is the point (-2,0)

Part 2) The solutions are the points (-2,0) and (0,2)

Step-by-step explanation:

Part 1

we have

$y=x+2$ -----> equation A

$y=0.5(x+2)$ ----> $y=0.5x+1$ ---> equation B

Solve the system by graphing

Remember that the solution of the system is the intersection point both graphs

using a graphing tool

The solution is the point (-2,0)

see the attached figure N 1

Part 2

we have

$y=x+2$ -----> equation A

$y=0.5(x+2)^2$ ----> equation B

Solve the system by graphing

Remember that the solution of the system are the intersection points both graphs

using a graphing tool

The solutions are the points (-2,0) and (0,2)

see the attached figure N 2

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If BC = 61 and AB = 82, what is the length of AC?

Nitella [24]

Answer:

54.8

Step-by-step explanation:

Pythagorean theorem is a^2+b^2=c^2.

Plug the given values in to get 61^2+b^2=82^2

61^2 is 3,721 and 82^2 is 6,724.

6,724-3,721 is 3,003.

The square root of 3,003, rounded to a reasonable value, is 54.8.

The length of AC is 54.8.

4 0

3 years ago

CAN UHELP PLEASE???? I NEED HELP AND I CAN TAKE THETEST ONLT ONE TIME !PLLLLZZZZZ!

marusya05 [52]

Well add them all up. : 10+15+8+7=40. There are 8 white balls out of 40. 8/40. Find the GCF of them both and simplify them.

8=8 so 8/8 = 1
40=8 so 40/8 = 5 so 1/5 is your answer.

B. is your answer.

Hoped I helped!

4 0

3 years ago

Read 2 more answers

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$