Find the value of x.

Solve the equation. Check the solution.<br> 1/2(2x-1)= -1/9 -7/9

Thepotemich [5.8K]

Answer:

hope this can help you (^^)

3 0

3 years ago

Lisa is 4 years older than three times her brother Sammy’s age.

Alina [70]

The correct choice is d

4 0

3 years ago

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PLEASE HELP!!! Pr-Cal Question 20 points

Brilliant_brown [7]

Answer:

3600

Step-by-step explanation:

perimeter is 120ft

Goal is to get the largest area.

so you can start with a list of possible solutions:

1*119 = 119ft

2*118 = 236ft

.

60*60 =3600ft

61*59 = 3599ft

.

7 0

3 years ago

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Find the measure of angle X in the triangle below.

Elan Coil [88]

Answer:

X = 89.92° or 90.08°

Step-by-step explanation:

The law of sines can be used to find the value of angle X:

sin(X)/26 = sin(67.38°)/24

sin(X) = (26/24)sin(67.38°) ≈ 0.99999901787

There are two values of X that have this sine:

X = arcsin(0.99999901787) ≈ 89.92°

X = 180° -arcsin(0.99999901787) ≈ 90.08°

There are two solutions: X = 89.92° or 90.08°.

_____

<em>Comment on the problem</em>

We suspect that the angle is supposed to be considered to be 90°. However, the given angle is reported to 2 decimal places, so we figure the requested angle should also be reported to 2 decimal places.

The lengths of the short side that correspond to the above angles are 10.03 and 9.97 units. If the short side were considered to be 10 units, the triangle would be a right triangle, and the larger acute angle would be ...

arcsin(24/26) ≈ 67.38014° . . . . rounds to 67.38°

This points up the difficulty of trying to use the Law of Sines on a triangle that is actually a right triangle.

3 0

3 years ago

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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

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

7 0

3 years ago