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bogdanovich [222]
3 years ago
12

Simplify 4 over quantity 2 plus 7 i

Mathematics
2 answers:
Marrrta [24]3 years ago
8 0
(8-28i)/53 get that a grade! :)))
deff fn [24]3 years ago
4 0

Answer:

(8-28i)/53

hope it helps!

Step-by-step explanation:

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Please Solve this 10+2y
Zigmanuir [339]
I believe the answer is 5y :)
6 0
3 years ago
I need help with this math worksheet.im not smart at all.
garik1379 [7]

Answer:

1) 11.581

2) 10.78

3) 14.8

4) 3.5

5) 16.28

6) .64

7) 23.3

8) 1.24

9) 6.53

10) 1.59

11) 8.79

12) 27.7919

Step-by-step explanation:

Ur smarter than u think. ^_^

7 0
2 years ago
Read 2 more answers
Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. B = 15° C = 113° b = 49
VMariaS [17]

Answer:

  A = 52°, a = 149.2, c = 174.3

Step-by-step explanation:

Technology is useful for this. Many graphing calculators can solve triangles for you. The attachment shows a phone app that does this, too.

___

The Law of Sines can give you the value of c, so you can choose the correct answer from those offered.

  c = sin(C)·b/sin(B) = sin(113°)·49/sin(15°) ≈ 174.271 ≈ 174.3 . . . . . third choice

8 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
7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%28fog%29%28%20-%205%29" id="TexFormula1" title="(fog)( - 5)" alt="(fog)( - 5)" align="absmidd
Xelga [282]
That should be the answer

8 0
3 years ago
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