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Zarrin [17]
3 years ago
9

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
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