Oops i meant what is 2 3/5 as a decimal?

Mathematics

2 answers:

Andrews [41]3 years ago

8 0

2 3/5 as a decimal is 2.6

MrMuchimi3 years ago

4 0

Answer:

Step-by-step explanation:

The answer is 4.6

You might be interested in

40 POINTS is (4,7) a solution of equations? Explain your answer 2x+4y=36 3x-4y=-6

dsp73

2x + 4y = 36
3x - 4y = -6
--------------add
5x = 30
x = 6

2x+4y=36
2(6)+4y=36
12 + 4y = 36
4y = 24
y = 6

answer
(6, 6) is a solution of equations
(4,7) is NOT a solution of equations

6 0

3 years ago

Read 2 more answers

what is the length of the diagonal of a non-regulation tennis court with length 20 feet and width 15 feet?

Semmy [17]

Answer:

25 feet

Step-by-step explanation:

Basically that non-regulation tennis court is a rectangle. You want to know the length of the diagonal. If you draw it on paper, you'll see that this then become 2 triangles... of which you have 2 sides, and are seeking the hypotenuse. So....

H² = A² + B²

H² = 20² + 15² = 400 + 225 = 625

H = 25 feet.

6 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

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