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3 years ago

Which probability indicates that an event is not likely to occur?

Mathematics

1 answer:

ira [324]3 years ago

7 0

The answer is 1 fourth because it's a 25% chance of happening and it's the lowest percentage , 1 is not the right answer because that is 100% so the answer is A (1/4

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The formula v = square root of 64h gives the velocity v in feet per second of an object that has fallen h feet. find the velocit

Trava [24]

Answer:

22.63

Step-by-step explanation:

v = √64 * h

= √64 * 8

= √64 * √8

= 8 * 2√2

= 16√2 = 22.63

7 0

3 years ago

Quadrilateral RUST has a vertex at R(2,3). What are the coordinates of R' after dilation by a scale factor of 3, centered at the

coldgirl [10]

Concept of this answer is (3,5) in the y(x)=-3

6 0

3 years ago

Read 2 more answers

SHOW UR WORK PLEASE IM LTERALLY CRYING

GalinKa [24]

Answer : 6 pieces
first you need to convert 3/4ths into 8th so you multiply
3/4 * 2 = 6/8
you can get 6 pieces of ribbon

4 0

3 years ago

What is the surface area of a cylindrical ring where the diameter of the cross section is 6.3 in and the center line has a lengt

Kruka [31]

The surface area of the cylindrical ring is given by
πdh
where d is the diameter and h is the height:
π•6.3•48 = 950.02
The answer is C. 950.02 in^2

4 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