Who else loves the book call me by your name

Connor jogs at a constant rate of 3.5 miles per hour. The table shows the distance traveled for different numbers of hours. Drag

allsm [11]

Answer:

1st and 3rd go in the 1st box

2nd and 4th go in the 2nd box

last one goes in the last box

Step-by-step explanation:

1,2,3,4,5 and time spent jogging go in independent quantity

3.5, 7, 10.5, 14, 17.5 and distance traveled go in dependent quantity

m=3.5h goes in neither

3 0

3 years ago

Simplify:<br> 5 1/12 - 2 3/4 =<br> Express your answer as a mixed number.

disa [49]

Answer:

Mixed Number Form: 2 1/3

Step-by-step explanation:

7 0

3 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

3 years ago

Can someone please help me asap?

pashok25 [27]

find the area the get the premiraterStep-by-step explanation:add

3 0

3 years ago

Use the distributive property to simplify the following expression: 5(x + 2)

OLga [1]

Answer:

5x + 10

Step-by-step explanation:

Each term in the parenthesis is multiplied by the 5 outside, that is

(5 × x) + (5 × 2) = 5x + 10

5 0

4 years ago