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2 years ago

Which Natural Numbers are not Integers?

Mathematics

1 answer:

Genrish500 [490]2 years ago

6 0

Answer:

Step-by-step explanation:

fraction are not integers

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How do you do number 15 thxxx

vampirchik [111]

The simpliest way to do this is to turn both fractions into decimals

7/10 goes to 0.7 and 2/5 goes to 0.4

then the equation would be 0.7x + 20 = 0.4

hope I've helped!

5 0

2 years ago

A person 5 feet tall cast a 11 foot shadow. A tree is casting a 27 foot shadow. How tall is the tree

Pavlova-9 [17]

Answer:

33?

Step-by-step explanation:

6 0

2 years ago

Solve the question on the image

Nataly [62]

Times the whole equation by 15 since the LCM of 3 and 5 is 15.
You get: 5p-3p-6= 60
Rearrange to get: 2p=66
p=33

5 0

1 year ago

At Westside Middle School, 10 percent, or 140 students, participate in soccer. How many students are in the school? Check all th

Alexeev081 [22]

Answer:

140 + 10 = 150, so 150 + 140 = 290 students. There are 140 + 10 = 150 students at Westside Middle School.

Step-by-step explanation:

5 0

2 years ago

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

2 years ago